Aboy 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. part b the ball's speed as it passes the window on its way down. part c the speed of impact on the ground.

A.25.096m

B.10m/s

V=22.189m/s

Explanation:

Part A The ball's maximum height above the ground. Part B The ball's speed as it passes the window on its way down. Part C The speed of impact on the ground.

newton's equation of motion

V^2= u^2+2as

a=-g since it is going against gravity

S=h s= distance travelled

h=height

V=final velocity

U=initial velocity

V=o

u=10m/s

0=100+2(-9.81)h

h=100/19.62

h=5.096m

The maximum height above ground level=

Height from thr window to the maximum the ball reached before coming downward+ height from the window to the ground level

H=h+h1

H=5.096+20

25.096m

2.part b

The final velocity as it reach its maximum height upward becomes zero and then the final velocity upward= the initial velocity when coming down

V=U =0

v^2=u^2+2ah

V^2=0+2(9.81)(5.096)

V=9.99917596605

V=10m/s

3. Speed of its impact on the ground

v^2=u^2+2ah

V^2=0+2(9.81)(25.096)

V^2=492.38352

V=22.189m/s

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.