Question

A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0-m-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 6.00 m/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

Answer 1

Answer:

a) 72.54°  and  It will take the woman 33.4m

Explanation:

The woman runs with a constant speed V1 = 6m/s

V2 = 20m/s

V1 = V2 cos θ

Cos θ= V1/V2= 6/20

Cos θ= 0.3

Cos^-1 0.3=72.54°

Using Range formular for projectile

R= (V2 Cos θ)/g (V2 Sin θ)^2 +sqrt(V2 Sin θ)^2 + 2gh)

R= (20cos72.54)(2Sin72.54+sqrt(20Sin72.54)^2 + 2×9.8×45

R=33.4m

b )

see the attached file