Question

# A ball is thrown vertically upward from a window that is 3.6 m above the ground....

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?

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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