Question

If the ball bounces straight up and returns to the floor 3.5 seconds after for striking it what was the balls greatest above the floor

Answer 1

Given,

The time, T = 3.5 s

The time the ball is required to go up is half of the time the ball bounces straight up and returns to the floor. It is the same for the ball going down.

`\begin{gathered} t=(T)/(2) \\ t=(3.5)/(2)s \end{gathered}`

`v=u+at`

Here, u is the initial velocity, v is the final velocity and a is the acceleration due to gravity calculated as g = 9.8 m/s^2.

Rearrange the equation,

`u=v-at`

Put t = 3.5/2,

`\begin{gathered} u=0-(-9.81m/s^2)*((3.5)/(2)s) \\ u=17.16\text{ }(m)/(s) \end{gathered}`

By the equation of the motion,

`\begin{gathered} s=ut+(1)/(2)at^2 \\ s=17.16*1.75+(1)/(2)*(-9.81)*(1.75)^2 \\ s=30.03-15.02 \\ s=15\text{ m} \end{gathered}`

Thus, the maximum height of the ball is 15 m.