Question

A ball is thrown vertically upward from a window that is 3.6 m above the ground. The ball's initial speed is 2.8 m/s and the acceleration due to gravity is 9.8 m/s^2.

a) What is the ball's speed when it hits the ground?

b) How long after the first ball is thrown should a second ball
be simply dropped from the same window so that both balls hit the
ground at the same time?

Answer #1

There are two ways to find the speed at the bottom.

First you could use the equations of motion

V^2 = Vi^2 + 2gh = 2.8^2 = 2 x 10 x 3.6 = 80 take the square root and you get 8.9 m/s.

Second we could use conservation of energy.

KE + PE at the top = KE at the bottom

0.5 mv^2 + mgh = 0.5 mV^2 the m's cancel out.

0.5 x 2.8^2 + 10 x 3.6 = 0.5 V^2

**V = 8.9
m/s**

B) To get the time that it takes to fall use T = (Vf - Vi) / A

T = (8.9 - - 2.8) / 10 = 1.2 seconds.

The one that is dropped d = 0.5 A T^2

3.6 = 5 T^2

**T = 0.8 seconds** so it should be dropped 0.4
seconds after to hit the ground at the same time.

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