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?
Question
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.