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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鈥檚 trajectory as viewed by (i) a person at rest on the ground and (ii) the runner.

Short Answer

Expert verified

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

(b) (i)

(b) (ii)

Step by step solution

01

Newton’s law

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

According to the newton鈥檚 laws of motion,

s=ut+12at2

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

For the vertical and horizontal motion, the velocity component will berole="math" localid="1663739048893" u蝉颈苍胃andcos respectively.

02

Given

Weight of ball =2.7Kg

Initial speed = 20m/s

Cliff height = 45m

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

03

(a) Angle at which the ball should be thrown

For the horizontal speed,

uh=ucoscos=6.00m/s20.00m/s=72.5

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

s=6.00m/s5.55ss=33.3m

04

(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.

05

(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.

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