Q64P - University Physics with Modern Physics

Question

A 2.7kKg ball is thrown upward with an initial speed of 20m/s from the edge of a high cliff. At the instant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 60m/s. The woman runs in a straight line on level ground. Ignore air resistance on the ball. (a) At what angle above the horizontal should the ball be thrown so that the runner will catch it just before it hits the ground, and how far does she run before she catches the ball? (b) Carefully sketch the ball’s trajectory as viewed by (i) a person at rest on the ground and (ii) the runner.

Step-by-Step Solution

Answer

(a) The angle should be $\theta =72.5°$ and the distance run by the woman s=33.3m.

 

(b) (i)

(b) (ii)

Step 1: Newton’s law

The velocity contains two components, one is horizontal component and another one is vertical. 

 

According to the newton’s laws of motion,

 

$s=ut+\frac{1}{2}a{t}^2$

 

Where, s, u, t, a, are displacement, initial velocity, time, and acceleration.

 

For the vertical and horizontal motion, the velocity component will be  $\mathrm{u} \mathrm{sin\theta } \mathrm{and} \cos \mathrm{\theta }$ respectively.

Step 2: Given

Weight of ball = $2.7\mathrm{Kg}$

 

Initial speed = 20m/s

 

Cliff height = 45m

 

Woman’s speed during running away from the base of cliff = 6.0m/s 

Step 3: (a) Angle at which the ball should be thrown

For the horizontal speed,

$$\begin{gathered} u_h=u\cos \theta \\ \cos \theta =\frac{6.00m/s}{20.00m/s} \\ \theta =72.5° \end{gathered}$$ 

 

The ball is in air for 5.55s so the horizontal distance run by her,

$$\begin{gathered} \mathrm{s}=\left(6.00\mathrm{m}/\mathrm{s}\right)\left(5.55\mathrm{s}\right) \\ \mathrm{s}=33.3\mathrm{m} \end{gathered}$$ 

Step 4: (b) (i) The ball’s trajectory as viewed by a person at rest on the ground

When the ball trajectory will be viewed by a person at the rest on the ground the ball moves in a parabolic curve.

Step 5: (b) (ii) The ball’s trajectory as viewed by the runner.

When the ball trajectory will be viewed by a the runner   the motion of ball will be only vertical.