Question

A ball is thrown from a ditch that is 12 feet below ground level. It had a speed of 64ft/s as it leaves his hand. what is the maximum height of the ball?

Answer 1

`\begin{gathered} \text{Max height is given as } \\ h=ut-(1)/(2)gt^2 \\ \text{given u = 64ft/s and g=32ft/s}^2 \\ h=64t-(1)/(2)(32)t^2 \\ h=64t-16t^2--------\text{eqn 1} \end{gathered}`
`\text{ speed = }(dh)/(dt)=\text{ 64-32t}`

At maximum height , final velocity = 0

Hence

0 = 64-32t

32t=64

t=64/32

t=2

substitute t= 2 into eqn--------------- 1

`\begin{gathered} h=64(2)-16(2^2) \\ h=128-64 \\ h=64ft \end{gathered}`

The maximum height covered from below the ground level (in the ditch) is 64ft but Above the ground level, we have 64ft - 12ft which is 52ft above ground level

3. Time taken before the ball hit the ground is obtained by finding the root of

`64t-16t^2=0`

=t (64-16t)=0

64 - 16t=0

16t = 64

t=64/16

t = 4s

Time taken before the ball hit the ground is 4s

4. Vertical velocity before hitting the ground is given as

`\begin{gathered} V\text{ = }\sqrt[]{2gh} \\ V=\text{ }\sqrt[]{2\text{ x 32 x64}} \\ V=32\sqrt[]{2}\text{ ft/s or 45.25ft/3} \end{gathered}`