A cannon ball is shot from ground level towards a target. Its initial velocity
is v_{0} = 125 m/s at an angle θ = 37 degrees with the horizontal.
Neglect air resistance.

(a) What are the horizontal and vertical components of the initial velocity?

(b) What is the maximum height of the cannonball? How long does it take to
reach this height?

(c) How long does it take to hit the ground? When it does so, what is its
horizontal distance from its starting point?

(d) What are its height and its horizontal displacement after it has been in
the air for 10 s?

What is its velocity (magnitude and direction) after it has been in the air for
10 s?

Solution:

- Concepts:

Kinematic equations, projectile motion - Reasoning:

We have motion with constant acceleration. - Details of the calculation:

(a) v_{0}= 125 m/s, θ = 37 degrees.

v_{x}(0) = v_{0}cosθ = 99.8 m/s ~100 m/s, v_{y}(0) = v_{0}sinθ = 75.2 m/s ~ 75 m/s.

(b) v_{y}(t_{max-height}) = v_{y}(0) - gt_{max-height}= 0, t_{maxheight}= 7.7 s.

v_{y}(t)^{2}= v_{y}(0)^{2}- 2gy, v_{y}(0)^{2}= 2gy_{max}, y_{max}= 287 m.

(c) Time to hit ground = 2*7.7s = 15.4 s. Distance = v_{0}cosθ*15.4 s = 1540 m.

(d) y(10 s) = 260 m, x(10 s) = 1000 m.

v_{x}(10 s) = 100 m/s, v_{y}(10 s) = -23 m/s, v = (v_{x}^{2}+ v_{y}^{2})^{½}= 102.6 m/s.

tanθ = v_{y}/v_{x}, θ ~ -13^{o}.

Two trucks are parked back to back in opposite directions on a straight, horizontal road. The trucks quickly accelerate simultaneously to 3.0 m/s in opposite directions and maintain these velocities. When the backs of the trucks are 20 meters apart, a boy in the back of one truck throws a stone at an angle of 40 degrees above the horizontal at the other truck. How fast must he throw, relative to the truck, if the stone is to just land in the back of the other truck?

Solution:

- Concepts:

Kinematic, projectile motion - Reasoning:

The stone's motion is motion in two dimensions with constant acceleration.

Put the origin of the coordinate system at the back of the parked trucks and let the boy be in the truck moving into the negative x direction. When each truck has moved 10 m, is speed is v = 3 m/s. Let v be the speed of the stone relative to the truck. - Details of the calculation:

The x-component of the stone's speed with respect to the ground is v_{x}= v cos40^{o}- 3 m/s.

The y-component of the stone's speed with respect to the ground is v_{y}= v sin40^{o}.

For the stone is to just land in the back of the other truck we need

v_{x}t = 20 m + (3 m/s)t. 0 = v_{y}t - ½gt^{2}, t = 2v_{y}/g.

Therefore

2v_{x}v_{y}/g = 20 m + (3 m/s) 2v_{y}/g.

v_{x}v_{y}= g 10 m + (3 m/s) v_{y}.

(v cos40^{o}- 3 m/s) v sin40^{o}= g 10 m + (3 m/s) v sin40^{o}.

v^{2 }cos40^{o }sin40^{o}= g 10 m + (6 m/s) v sin40^{o}.

v^{2}= 199 (m/s)^{2}+ 7.83 (m/s) v.

v = 18.56 m/s.

A player tossed a ball at some angle relative to the horizon. The maximum speed of the ball during the flight was 12 m/sec and the minimum speed was 6 m/sec. What was the maximum height of the ball during the flight? Please neglect air resistance.

Solution:

- Concepts:

Kinematics, projectile motion - Reasoning:

The ball executes projectile motion. - Details of the calculation:

v_{x }= v_{0}cosθ_{0 }= constant, x = v_{0}cosθ_{0}t,

v_{y }= v_{0}sinθ_{0 }- gt, y = v_{0}sinθ_{0}t - ½gt^{2}.

Since v_{x }= constant and v_{y}= 0 when the ball reaches its maximum elevation,

v_{max}= v_{0}= 12 m/s and v_{min}= v_{0}cosθ_{0}= 6 m/s. Therefore cosθ_{0}= ½, θ_{0}= 60^{o}.

t_{max_height }= v_{y0}/g = v_{0}sinθ_{0}/g. y_{max}= ½(v_{0}sinθ_{0})^{2}/g = 5.5 m.

A rock is launched from the ground level at a^{ }
speed v directed at an angle θ with the horizontal.^{ }It is
noticed that some (unknown) time t after^{ }the launch, the distance
between the rock and the launch^{ }point begins to decrease.

(a)
Find the smallest launch angle θ consistent with this observation.

(b)
Find t, neglecting the air resistance.

Solution:

- Concepts:

Kinematics, projectile motion - Reasoning:

The rock is a projectile.

For projectile motion we have x = v cosθ_{ }t, y = v_{ }sinθ_{ }t - ½gt^{2}. - Details of the calculation:

(a) The distance between the rock and the launch^{ }point is D = (x^{2}+ y^{2})^{½}= (v^{2}t^{2}+ g^{2}t^{4}/4 - vgt^{3}sinθ)^{½}.

For 0 < θ < 90^{o}and v > 0 the distance D increases initially. To find extrema for D we set

dD^{2}/dt = t(g^{2}t^{2}- 3vgt sinθ + 2v^{2}) = 0.

t^{2}- 3(v/g) sinθ t + 2v^{2}/g^{2}= 0, t = [3v/(2g)][sinθ ± (sin^{2}θ - 8/9)^{½}].

If sin^{2}θ > 8/9 or θ > 70.5287^{o}, then two extrema exist.

The figure below shows a plot of D versus t for v = 10 m/s and a launch angle smaller, equal, or greater than the critical angle θ > 70.5287^{o}.

The smallest launch angle θ consistent with the observation is just greater than 70.5287^{o}.

(b) t = t_{1}= [3v/(2g)][sinθ - (sin^{2}θ - 8/9)^{½}] is the maximum we are looking for. For t > t_{1}D decrease and for

t > t_{2 }= [3v/(2g)][sinθ + (sin^{2}θ - 8/9)^{½}]_{ }it increases again.

The rock hits the ground at t = 2vsinθ/g.

Two points, A and B, are located on the ground a certain distance d apart.
Two rocks are launched simultaneously from points A and B, with equal speeds but
at different angles. Each rock lands at the launch point of the other.
Knowing that one of the rocks is launched at an angle θ_{0} > 45^{o},
find the minimum distance between the rocks during the flight in terms of d and
θ_{0}?

Solution:

- Concepts:

Kinematics, projectile motion - Reasoning:

We have to find the position of both rocks as a function of time.

The range of a projectile launched with speed v_{0}at an angle θ_{0}with respect to the horizontal x-axis is d = v_{0}^{2}sin(2θ_{0})/g.

For two projectile launched with the same speed v_{0}to have the same range they have to be launched at angles θ_{±}= π/4 ± θ. - Details of the calculation:

Assume rock 1 is thrown from point A with speed v_{0 }and launch angle

θ_{+}= θ_{0}= π/4 + θ towards point B.

Let the origin of the coordinate system be at point A, let the x-axis point towards B and let the y-axis point upward.

For rock 1 we have

x_{1}(t) = v_{0}cos(π/4 + θ)t, y_{1}(t) = v_{0}sin(π/4 + θ)t - ½gt^{2}.

Assume rock 2 is thrown from point with speed v_{0 }and launch angle θ_{-}= π/4 - θ towards point A.

For rock 2 we have

x_{2}(t) = d - v_{0}cos(π/4 - θ)t, y_{2}(t) = v_{0}sin(π/4 - θ)t - ½gt^{2}.

Let the distance between the rocks be L(t).

L^{2}(t) = (x_{2}(t) - x_{1}(t))^{2}+ (y_{2}(t) - y_{1}(t))^{2}

= v_{0}^{2}t^{2}(d/(v_{0}t) - cos(π/4 - θ) - cos(π/4 + θ))^{2}+ v_{0}^{2}t^{2}(sin(π/4 - θ) - sin(π/4 + θ))^{2}

= v_{0}^{2}t^{2}(d/(v_{0}t) - cos(π/4)cos(θ) - sin(π/4)sin(θ) - cos(π/4)cos(θ) + sin(π/4)sin(θ))^{2}

+ v_{0}^{2}t^{2}(sin(π/4)cos(θ) - cos(π/4)sin(θ) - sin(π/4)cos(θ) - cos(π/4)sin(θ))^{2}

= 4v_{0}^{2}t^{2}(d/(2v_{0}t) - cos(π/4)cos(θ))^{2}+ 4v_{0}^{2}t^{2}cos^{2}(π/4)sin^{2}(θ)

= d^{2}+ 4v_{0}^{2}t^{2}cos^{2}(π/4)(cos^{2}(θ) + sin^{2}(θ)) - 4v_{0}^{2}t^{2}dcos(π/4)cos(θ)/(v_{0}t)

= d^{2}+ 2v_{0}^{2}t^{2}- 4v_{0}tdcos(π/4)cos(θ).

To find the time at which L^{2}(t) has a minimum, set

dL^{2}(t)/dt = 4v_{0}^{2}t - 4v_{0}dcos(π/4)cos(θ) = 0.

t_{min}= dcos(π/4)cos(θ)/v_{0}.

L^{2}(t_{min}) = d^{2}+ d^{2}cos^{2}(θ) - 2d^{2}cos^{2}(θ) = d^{2}- d^{2}cos^{2}(θ).

L_{min}= dsin(θ) = dsin(θ_{0}- π/4) is the minimum distance between the rocks.

A catapult set on the ground can launch a rock a maximum horizontal distance
L. What would be the maximum horizontal launch distance if the catapult is
set on a platform moving forward with constant speed equal to the launch speed
of the rock?

Neglect the air resistance and assume that the rock is launched
from the ground level in both cases.

Solution:

- Concepts:

Kinematic equations, projectile motion - Reasoning:

We have motion with constant acceleration. - Details of the calculation:

Let θ_{0}be the launch angle with respect to the horizontal. (v_{x}= v cosθ_{0}, v_{y}= v sinθ_{0})

The range of a projectile is R = (v_{0}^{2}sin2θ_{0})/g.

For the stationary catapult: R_{max}= v_{0}^{2}/g = L, v_{0}= (gL)^{1/2}. (θ_{0}= 45^{o})

For the moving catapult: R = (v_{0}^{2}sin2θ_{0})/g + v_{0}t.

t = 2v_{y}/g = 2v_{0}sinθ_{0}/g.

R = (v_{0}^{2}sin2θ_{0})/g + 2v_{0}^{2}sinθ_{0}/g,

dR/dθ_{0}= 0 --> cos2θ_{0}= -cosθ_{0. }θ_{0}= 60^{0}.

R = (v_{0}^{2}/g)[sin2θ_{0}+ 2sinθ_{0}]

= L 3√3/2.

Consider a lawn sprayer consisting
of a spherical cap (α_{0}
= 45^{o}) provided with a large number of equal holes through which
water is ejected with speed v_{0}. The lawn is not evenly sprayed
if the holes are evenly spaced. How must the number of holes per unit
area, r(α),
be chosen to achieve uniform spraying of a circular area? Assume the
radius of the sprinkler cap is very much less than the radius of the area to be
sprayed, and the surface of the cap is at the level of the lawn.

Solution:

- Concepts:

Kinematics, projectile motion - Reasoning:

The drops are projectiles. They execute motion in more than one dimension motion with constant acceleration. This acceleration is directed downward and has magnitude g.

A drop sprayed from an angle a travels a distance R(α). We want to find this range R in term of α.

We use

v_{y0 }= v_{0}cosα, v_{x0 }= v_{0}sinα,

x = x_{0}+ v_{x0}t, y = y_{0 }+ v_{y0}t - ½gt^{2},

and substitute

0 = 0 + v_{0}(cosα)t - (½gt^{2}),^{ }t = 2v_{0}cosα /g

to solve for

R = x(t) = v_{0}(sinα)t = v_{0}(sinα)2v_{0}(cosα)/g = v_{0}^{2}(sin2α )/g. - Details of the calculation:

Drops sprayed from any angles between α and α + dα fall on an area 2πRdR between R and R + dR.

The number of drops sprayed from angles between α and α + dα is proportional to ρ(α)2πr^{2}_{cap}sin α dα.

We want

[ρ(α)2πr^{2}_{cap}sinα dα]/[2πRdR] = const,

or

(ρ(α)sinα)/(RdR/dα) = const.

dR/dα = 2v_{0}^{2}(cos2α)/g.

Therefore

ρ(α )(sinα)/(cos2α sin2α) = const,

or

ρ(α ) is proportional to sin(4α) /sinα.

A particle is launched with velocity **v** = 2 ft/s
**j** along the ridge
of
a roof, but the equilibriums is unstable and it immediately starts
accelerating down the right side of the roof. Using the coordinate

system in the figure and neglecting friction, what are the particle's
x- and y-coordinates when it hits the ground?

Use the following measurements: |

Roof Rise: z_{R} = 4 ft; Roof Width: x_{R}= 24 ft; Roof Length:
y_{R} = 24 ft;

Wall Height: z_{W} = 8 ft; g = 32 ft/s^{2}.

Solution:

- Concepts:

Kinematics, motion with piecewise constant acceleration, energy conservation - Reasoning:

The forces acting on the particle are gravity and the normal force of the roof. The motion in the xz plane is independent of the motion with constant v_{y}in the y-direction. - Details of the calculation:

Consider only the motion in the xz plane: mg∆h = ½mv^{2}. (v^{2}= v_{x}^{2}+ v_{y}^{2})

As long as the particle is in contact the roof: a = gsinθ. sinθ = 4/(160)^{½}= 0.316.

For the motion to the edge of the roof: v_{1}= (2g z_{R})^{½}= 16 ft/s v_{1}= gsinθ t_{1}, t_{1}= 1.58 s.

[Check that the particle stays on the roof during the time interval t_{1}.

During time t_{1}, the particle travels a distance d = 2 ft/s *t_{1}= 3.16 ft in the y direction, which is less than y_{R}. The particle stays on the roof. At the edge of the roof the particle's velocity is

**v**= g sinθ cosθ t_{1}**i**- g sinθ sinθ t_{1}**k**+ 2 ft/s**j**. =15.18 ft/s**i**- 5.06 ft/s**k**+ 2 ft/s**j**.]

The particle contacts the floor after an additional time t^{2}. 8ft - v_{1z}t_{2}- ½gt_{2}^{2}= 0 t_{2}= 0.568 s.

The particle's x and y coordinates at time t_{2}are x = 12 ft + g sinθ cosθ t_{1}t_{2}= 20.62 ft,

y = 3.16 ft + 2 ft/s (t_{2 }- t_{1}) = 4.3 ft.

In a carnival game, you have to throw a ball with speed v_{0 }at an
angle θ in order to hit a target on the other side of the platform, located a
distance h away. The platform is inclined at an angle φ. Find the angle θ in
terms of the other variables.

Solution:

- Concepts:

Kinematics, projectile motion - Reasoning:

Orient the x and y axes of your coordinate system as shown.

The net force acting on the ball in a direction perpendicular

to the xy-plane is zero.

In the x-y plane a_{x}= 0, a_{y}= -gsinφ = -g'. - Details of the calculation:

The range of the ball is h = (v_{0}^{2}sin2θ)/g'.

sin2θ = h gsinφ/v_{0}^{2}, θ = ½ sin^{-1}(h gsinφ/v_{0}^{2}).

A tree trunk lies on the ground. The trunk has a shape of a cylinder with
radius R. A flea attempts to jump over the trunk. What is the minimum initial
speed v that enables the flea to reach the other side? Assume that the flea is
intelligent enough to select the optimal take-off point on the ground.

Consider two cases.

(a) The flea is allowed to slide on the frictionless trunk.

(b) The flea is not allowed to slide on the trunk and must clear the trunk.

Solution:

- Concepts:

Kinematics, projectile motion - Reasoning:

(a) Imagine the flea sitting on top of the trunk. This is an unstable equilibrium and the slightest perturbation will cause it to slide off. Its kinetic energy when reaching the ground will be

½mv^{2}= 2mgR. The motion may be reversed.

(b) The flea's trajectory must be a parabola as shown.

- Details of the calculation:

(a) he flea must jump with a speed greater than v = 2(gR)^{½}.

(b) Let the parabolic arc of the flea be tangent to the trunk at an angle θ from the top of the trunk.

The velocity of the flea at this point is**v**_{θ}= (v_{θ}cosθ, v_{θ}sinθ).

The conditions for the parabola to reach its maximum over the center of trunk are

v_{θ}cosθ t = r sinθ, v_{θ}sinθ = gt for some t. (**v**_{θy}must be zero at the top.)

These give v_{θ}^{2}= rg/cosθ.

Let v be the speed with which the flea jumps from the ground. Then the energy at the ground is

½ mv^{2}= mg(r + r cosθ) + ½mrg/cosθ = mgr(1 + cosθ + 1/(2cosθ)).

Minimizing this function of cosθ yields cosθ = 1/√2, θ = 45^{o}.

The minimum initial speed is v^{2}= 2gr(1 + √2), v = 2.197 (gr)^{½}.

A marble bounces down stairs in a regular manner, hitting each step at the same place and bouncing the same height above each step.

The stair height equals its depth (tread = rise) and the
coefficient of restitution ε is given. Find the necessary
horizontal velocity and bounce height.

(The coefficient of restitution is defined as ε = -v_{f}/v_{i},
where v_{f} and v_{i} are the vertical velocities just after and
before the bounce. respectively).

Solution:

- Concepts:

Kinematics, collisions, motion with constant acceleration - Reasoning:

The horizontal component of the velocity of the marble, v_{h}, is constant. Let v_{i}and v_{f}denote the magnitudes of the vertical components of the velocity of the marble just before and just after a bounce, respectively. Since the conditions at each step are exactly the same, v_{i}and v_{f}are also constants. The relationship between the magnitudes of**v**_{1}= v_{h}**i**- v_{i}**j**and**v**_{2}= v_{h}**i**+ v_{f}**j**is found from the conservation of mechanical energy in a conservative field. ½mv_{2}^{2}+ mgl = ½mv_{1}^{2}. - Details of the calculation:

v_{2}^{2}+ 2gl = v_{1}^{2}. v_{i}^{2}- 2gl = v_{f}^{2}= ε^{2}v_{i}^{2}. v_{i}^{2}= 2gl/(1 - ε^{2}).

What is the time between bounces?

Starting with a vertical velocity v_{f}, the time to reach the point where the vertical velocity is zero is t_{1}= v_{f}/g.

The time to accelerate downward from this point and reach a vertical velocity -v_{i}is t_{2}= |v_{i}|/g.

The total time between bounces is t = t_{1}+ t_{2}= (v_{f}+ |v_{i}|)/g = l/v_{h}, since during this time the marble has to cover a horizontal distance l.

v_{h}= gl/(v_{f}+ |v_{i}|) = gl/[(1 + ε)|v_{i}|] = (gl/2)^{½}[(1 - ε)/(1 + ε)]^{½}.

The bounce height h is found from conservation of energy. ½mv_{f}^{2}= mgh, h = v_{f}^{2}/(2g) = ε^{2}l/(1 - ε^{2}).

(a) An astronaut on a strange
planet finds that she can jump a maximum horizontal distance of 15 m if her
initial speed is 3 m/s. What is the free-fall acceleration on the planet?

(b) How much work is required to raise a 100 g block to a
height of 200 cm and simultaneously give it a velocity of 300 cm/sec?

An plane, inclined at θ = 20^{o}, touches a wall as shown in the
picture. You drop a small, perfectly elastic ball from a height h = 1.5 m
onto the onto the plane. The ball falls from rest. You do not move
your hand. At what distance d from the wall do you have to drop it so that
it bounces back into your hand?

A wheel of radius b is rolling along a muddy road with a speed v. Particles
of mud attached to the wheel are being continuously thrown off from all points
of the wheel. If v^{2 }> 2bg, where g is the acceleration of gravity,
find the maximum height above the road attained by the mud, H = H(b,v,g).