Question

A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a horizontal distance of 17 m, it has dropped 55 cm from its original height when it left the racquet. What was the initial speed, in m/s, of the tennis ball

Answer 1

The value is
`u_x = 56.1 \ m/s`

Step-by-step explanation:

From the question we are told that

The horizontal distance traveled is
`h = 17 \ m`

The decrease in height is

Considering the vertical velocity

Generally from kinematic equation we have that

`v_y ^2 = u_y ^2 + 2* g * s`

Here the initial vertical (
`u_y`) velocity is zero

=>
`v_y ^2 = 0 ^2 + 2 * 9.8 * 0.55`

=>
`v_y = 2.97 \ m`

Generally the time to drop to this height is mathematically represented as

`t = (v_y )/(g )`

=>
`t = ( 2.97 )/( 9.8 )`

=>
`t = 0.3031 \ s`

Generally given that time taken to drop to the obtained height is equal to the time taken to reach the horizontal distance , then

`u_x = (h)/(t)`

Here
`u_x` is the initial horizontal velocity

`u_x = ( 17 )/( 0.3031 )`

=>
`u_x = 56.1 \ m/s`