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, and air resistance acting on the ball can be ignored.How far does the woman run before she catches the ball?

Answer 1

Answer:

The distance traveled by the woman is 34.1m

Explanation:

Given

The initial height of the cliff

yo = 45m final, positition y = 0m bottom of the cliff

y = yo + ut -1/2gt²

u = 20.0m/s initial speed

g = 9.80m/s²

0 = 45.0 + 20×t –1/2×9.8×t²

0 = 45 +20t –4.9t²

Solving quadratically or by using a calculator,

t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s

So this is the total time it takes for the ball to reach the ground from the height it was thrown.

The distance traveled by the woman is

s = vt

Given the speed of the woman v = 6.00m/s

Therefore

s = 6.00×5.69 = 34.14m

Approximately 34.1m to 3 significant figures.