Question

Joanna drops a box from the top of a large open window. The height of the window is 5.00 m. (Asuume the magnitude of the gravitational acceleration is g = 10 m/s^2.)

A) If Joanna is timing the box drop, how long will it take to reach the bottom of the window?

B) Her friend is standing 15.00m below the bottom of the window and is going to catch the falling box. How fast will the box be traveling at the moment before Joanna's friend catches it?

Answer 1

initial position yo = 0

final position y = -5 m

acceleration ay = -10 m/s^2

initial velocity voy = 0

(A)

from equation of motion

y - yo = voy*t + (1/2)*ay*t^2

-5 - 0 = 0*t - (1/2)*10*t^2

t = 1 s

it will take 1 s to reach the ground

=================================

B)

initial position yo = 0

final position y = -20 m

acceleration ay = -10 m/s^2

initial velocity voy = 0

(A)

from equation of motion

vy^2 - voy^2 = 2*ay*(y-y0)

vy^2 - 0 = -2*10*(-20-0)

vy = 20 m/s

the box is travelling at a speed of 20 m/s