a tennis ball is struck such that it leaves the racket horizontally with a speed of 28.0 m/s. the ball hits the court at a horizontal distance of 19.6 m from the racket. what is the height of the tennis ball when it leaves the racket?

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26-10-2022
Physics

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a tennis ball is struck such that it leaves the racket horizontally with a speed of 28.0 m/s. the ball hits the court at a horizontal distance of 19.6 m from the racket. what is the height of the tennis ball when it leaves the racket?

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qwterabite qwterabite · Answer 1 · 2022-10-29T04:14:24-04:00

the height of the tennis ball when it leaves the racket is 2.40 m

We are given the initial velocity of the ball 28.0 m/s and the final velocity will be 0 m/s, the distance of the court 19.6 m, t is the time of the flight and h is the height at which the tennis ball was

Since we know,

speed = distance/ time

=>time = distance /speed

=>t = 19.6 m/l 28.0 =0.7s

the ball took 0.7s to cover the distance of 19.6 m

Since using the equation of motion:

s=ut+1/2at², here is the acceleration due to gravity which is 9.8 ms⁻²

u is the initial velocity, s the distance vertically of the height, and t is the time taken

=0+1/2 9.8(0.7)^2

= 2.40 m

To know more about the equation of motion refer to the linkhttps://brainly.com/question/15080570?referrer=searchResults.

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