Question

A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he releases it. Use conversation of energy to find

Part A

The ball's maximum height above the ground.
Express your answer using two significant figures.

hmax = ___ m
Part B

The ball's speed as it passes the window on its way down.
Express your answer using two significant figures.

vwindow = ___ m/s
Part C

The speed of impact on the ground.
Express your answer using two significant figures.

vimpact = ____ m/s

Answer 1

25 m

9.9 m/s

22 m/s

Step-by-step explanation:

m = Mass of ball

v = Velocity

g = Acceleration due to gravity = 9.81 m/s²

Applying conservation of energy

`mgh=(1)/(2)mv^2\\\Rightarrow h=(v^2)/(g)\\\Rightarrow h=(10^2)/(2* 9.81)\\\Rightarrow h=5.09683\ m`

The height above the ground is 5.09683+20 = 25.09683 m = 25 m

`mgh=(1)/(2)mv^2\\\Rightarrow v=√(2gh)\\\Rightarrow v=√(2* 9.81* 5)\\\Rightarrow v=9.9\ m/s`

The ball's speed as it passes the window on its way down is 9.9 m/s

`mgh=(1)/(2)mv^2\\\Rightarrow v=√(2gh)\\\Rightarrow v=√(2* 9.81* 25)\\\Rightarrow v=22.14723\ m/s`

The speed of impact on the ground is 22 m/s