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}.

Find the minimum rotational frequency (i.e. the number of rotations per second, not the angular frequency) required to rotate a bucket containing 4 liters of water in a circular path in a vertical plane at an arm's length (about 70 cm) without spilling water.

Solution:

- Concepts:

Centripetal acceleration - Reasoning:

Assume constant ω. Then on top of the circular path the centripetal acceleration needed to keep the water moving in a circle must be greater than or equal to the gravitational acceleration g.

Limiting case: v^{2}/R = g, v = (gR)^{½}. - Details of the calculation:

f = v/(2πR) = (g/R)^{½}/(2π) = ((9.8 m/s^{2})/(0.7 m))^{½}/(2π) = 0.6/s.

A string with 10 beads (bead #0 to #9) attached to it at equal intervals d_{0}
is dropped from a vertical height h. (Bead #0 is at height h when the string is
released.) The sound of each bead is heard when it hits the ground.

(a) Find t_{n}, the time the nth bead hits the ground and evaluate t_{n}/t_{0}.
Does the time interval between sounds increase or decrease a the beads hit the
ground?

(b) Assume we want equally spaced sounds. If the spacing between bead 0 and
bead 1 is d_{0}, what should be the spacing d_{n} between bead n
and bead n + 1?

Solution:

- Concepts:
Kinematics - Reasoning:

All beads have the same acceleration g, downward. The beads in the upper parts of the chain accelerate for a longer time before hitting the ground and therefore have higher speeds when hitting the ground. Each successive bead moves faster through the last distance d_{0}, and we therefore expect the time interval between sounds to decrease. If we want equal time intervals between sounds, the distance between the beads has to increase with height. - Details of the calculation:

(a) When the chain is released, the lowest bead (#0) is at height h

and the highest bead (#9) is at height h + 9*d.

The time it takes an object to fall a distance y is t = (2gy)^{½}(from y = ½gt^{2}).

So t_{n}= (2g)^{½}(h + nd_{0})^{½}(n = 0 through 9) is the time when the nth bead hits the ground.

t_{n}= (2gh)^{½}(1 + nd_{0}/h)^{½}, t_{n}/t_{0}= (1 + nd_{0}/h)^{½}.

t_{n}increases proportional to the square root of n, not linearly with n, which would be required for equally spaced sounds. The spacing between sounds decreases.

(b) If we want equally spaced sounds we need a variable spacing between beads.

We want t_{n}= t_{0}(1 + y_{n}/h)^{½}= t_{0}+ nT, where T is the time interval we choose.

We need 1 + y_{n}/h = (1 + nT/t_{0})^{2}, y_{n}= h 2n T/t_{0}+ h n^{2}T^{2}/t_{0}^{2}.

The spacing between bead n + 1 and bead n is d_{n}= y_{n+1}– y_{n}= n2hT^{2}/t_{0}^{2}+ hT^{2}/t_{0}^{2}+ h2T/t_{0}.

The spacing between bead 0 and bead 1 is hT^{2}/t_{0}^{2}+ 2hT/t_{0}= d_{0}, therefore d_{n}= d_{0}+ n 2hT^{2}/t_{0}^{2}, and T/t_{0}= 1 + (1 + d_{0}/h)^{½}.

The spacing between the beads increases linearly with n.

A swimmer wants to cross a river,
from point A to point B, as shown in the
figure. The distance d_{1} (from
A to C) is 200 m, the distance d_{2}
(from C to B) is 150 m, and the speed v_{r}
of the current in the river is 5 km/hour. Suppose that the swimmer’s velocity
relative to the water makes an angle of
θ = 45° with the line from A to C, as
indicated in the figure.

To swim directly from A to B, what speed u_{s}, relative to the water,
should the swimmer have?

Solution:

- Concepts:

Kinematics - Reasoning:

In the frame of the river bank, the time it takes the swimmer to move 200 m in the y direction (d_{1}) has to be the same time taken to go 150m in the x direction (d_{2}). - Details of the calculation:

d_{2}= ∆x = v_{x}t = (-u_{s}/√2 + v_{r})t, d_{1}= ∆y = v_{y}t = (u_{s}/√2 )t.

Here v_{x}and v_{y}are the swimmers velocity components relative to the river bank.

d_{2}/d_{1}= -1 + √2v_{r}/u_{s}. u_{s}= √2v_{r}/(d_{2}/d_{1}+ 1)

d_{2}/d_{1}= 3/4, u_{s}= 0.808*(5 km/hr) = 4.04 km/h = 1.12 m/s.