Question

Mary spots Bill approaching the dorm at a constant rate of 2 m/s on the walkway that passes directly beneath her window, 17 m above the ground. When Bill is 120 m away from the point below her window she decided to drop an apple down to him.

A.) How long should Mary wait to drop the apple if Bill is to catch it 1.75 m above the ground, and without either speeding up or slowing down?

B.) How far from directly below the window is Bill when Mary releases the apple?

Answer 1

Given that Mary is standing 17 m above the vertical distance from bill and bill will catch apple 1.75 above the ground, So Vertical distance traveled by apple = 17 - 1.75 = 15.25 m

Now Using 2nd kinematic equation, time taken by apple to reach at above vertical distance

h = U*t + (1/2)*a*t^2

15.25 = 0*t + (1/2)*9.81*t^2

t = sqrt (2*15.25/9.81) = 1.763 sec

Now in this time horizontal distance traveled by bill will be:

Constant speed of Bill = 2 m/sec

distance traveled by bill in 1.763 sec will be:

distance = speed*time = (2 m/sec)*(1.763 sec) = 3.526 m

So Mary should drop the apple when bill is horizontally 3.526 m away from the spot

So time for which Mary have to wait will be:

Initial Horizontal distance at which bill will catch apple = 120 m

distance at which apple starts falling = 3.526 m

wait time for Mary = remaining distance/speed of bill

wait time for Mary = (120 - 3.526)/2 = 58.237 sec (Ans of part A)

As Solved above

Bill should be at a horizontal distance 3.536 m from window when apple is dropped.

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