Question

Rodney hits a 0.060 kg tennis ball straight up with a speed of 15 m/s from a height of 2.2 m above the ground. How fast is the ball going when it is 9.5 m above the ground?

Answer 1

Step-by-step explanation:

From the question we are told that

The mass of the tennis ball is
`m = 0.060 \ kg`

The speed is
`u = 15 \ m/s`

The initial height of the ball is
`h_o = 2.2 \ m`

The height considered is
`h = 9.5 \ m`

Generally the time taken to reach the considered height is mathematically evaluated from the equation of motion as follows

`h = h_o + ut - (1)/(2) gt^2`

=>
`9.5 = 2.2 + 15t - (1)/(2) * 9.8 t^2`

=>
`7.3 = 15t - 4.9t^2`

=>
`4.9t^2 -15t + 7.3`

solving the above equation using quadratic formula

`t= 0.6070 \ s` and
`t_1 = 2.454 \ s`

Generally from kinematic equation we have that

`v = u - gt`

=>
`v= 15 - 9.8 * 0.6070`

=>
`v= 9.0514 \ m/s`

Note if we had used the second values of time the velocity at 9.5 \ m would have been negative which is not correct given that at maximum height velocity is zero