Physics plz help one more time?

Question

A tenis ball is thrown straight up with an initial speed of 22.5 m/s. it is caught at the same distance above the ground

a) how high did the ball rise?

b) How long does the ball remain in the air?

plz teach

Answer 1

?it is caught at the same distance above the ground?

22.5 m/s this is a unit of speed. You need

to correct the question if you mean 22.5 meters

BUT:

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if caught at 22.5 meters how high******

v= velocity when caught

1/2m(vi)^2-mgh=1/2m(v^2)

solving for v

v=sqr(vi^2-2*g*h)

=sqr((22.5)^2-2*9.8*22.5)

= 8.08 m/s when caught

change in speed = acceleration * time

(22.5-8.08)=9.8*t

t= (22.5 - 8.08) / 9.8 = 1.47 seconds how long******

check:

s=(22.5 * 1.47) - (0.5 * 9.8 * (1.47^2))

= 22.5 meters

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assuming you mean caught at max-height.

initial speed 22.5 m/s

time in air up:

final velocity=0 "gravity robbed all the 22.5m/s"

0=22.5-g*t

solve for t:

t=-22.5/-9.8 = 2.3 seconds up

How high...

s=Vi*t - (1/2)*g*t^2

(22.5 * 2.3) - (0.5 * 9.8 * (2.3^2)) = 25.83 meters (answer)

how long....

time up---> it was caught

2.3 (answer)