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## a tennis ball is thrown straight up at a speed of 40m/s and caught at the same level. calculate rhe maximum height reached by the ball

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## Answers ( )

Answer:81.6 m

Explanation:Answer: 81.6 m.

The time it takes gravity to slow 40 m/s to zero when it teaches maximum height is

-v(initial) / -g = t

-40 m/s / -9.8 m/s^2 = 4.08 s

The height reached is the average velocity times this time 4.08 s, with v(avg) = [v(initial) + v(final)] / 2 with v(final) = 0. v(avg) = v(initial) / 2 = 40 m/s / 2 = 20 m/s.

So the distance d of maximum height is

d = v(avg)•t

d = 20 m/s • 4.08 s = 81.6 m.

Answer:80m

Explanation:time taken for stone to reach max-height is t=final velocity – intial velocity / acceleration due to gravity

t=v-u/g

where:

v=0 , u=40m/s ,g=-10m/so

: t=0-40/-10

t=4secs

: max-height reach=ut + 0.5gt²

=(40)(4)+(0.5)(-10)(4)I

=160-5(16)

=160-80

=

80m