Question

You toss a tennis ball straight upward. At the moment it leaves your hand it is at a height of 1.3 m above the ground, and it is moving at a speed of 8.8 m/s.(c) When the tennis ball is at a height of 2.6 m above the ground, what is its speed?

Answer 1

`\begin{equation*} 7.21\text{ m/s} \end{equation*}`

Step-by-step explanation

Parameters given:

Initial height = 1.3 m

Initial speed = 8.8 m/s

Final height = 2.6 m

To find the speed of the tennis ball at a height of 2.6 m above the ground, we have to apply one of Newton's equations of motion:

`v^2=u^2-2gs`

where v = final speed

u = initial speed = 8.8 m/s

g = acceleration due to gravity = 9.8 m/s^2

s = distance traveled by the ball

Therefore, the speed of the ball at a height of 2.6 m (final speed) is:

`\begin{gathered} v^2=8.8^2-(2*9.8*(2.6-1.3)) \\ \\ v^2=8.8^2-(2*9.8*1.3) \\ \\ v^2=77.44-25.48=51.96 \\ \\ v=√(51.96) \\ \\ v=7.21\text{ m/s} \end{gathered}`

That is the answer.