Question

A ball is thrown up and is caught by the player after 4 seconds. How high did it go and with what velocity was it thrown? How far below the ball will be from its highest point after 3 seconds from start?

Answer 1

In case of upward motion of the ball from A to B

Initial velocity, u =?

Final velocity, v =$0$ (at maximum height)

Time taken by the ball to reach the highest point = $2s$ ( time of ascent = time of descent)

Acceleration due to gravity g =$-9.8m{s}^{-2}$ ( upward motion)

Finding the initial velocity of the ball

Using the first equation of motion,$v=u+gt$

$$\begin{gathered} v=u-gt \\ 0=u-9.8\times 2 \\ u=19.6m/s \end{gathered}$$

The initial velocity of the ball is $19.6m/s$.

Finding the maximum height (h) attained by the ball

Using the second equation of motion, $h=ut+\frac{1}{2}g{t}^2$

$h=19.6\times 2-\frac{1}{2}\times 9.8\times {\left(2\right)}^2$

$$\begin{gathered} h=39.2-19.6 \\ h=19.6m \end{gathered}$$

Let the ball is at C after t = 3 sec

Consider motion from A to C

$$\begin{gathered} u=19.6m{s}^{-1},t=3s,g=-9.8m{s}^{-2},s=h\text{'} \\ s=ut+\frac{1}{2}g{t}^2 \\ h\text{'}=19.6\times 3-\frac{1}{2}\times 9.8\times {\left(3\right)}^2 \\ h\text{'}=58.8-44.1=14.7m \end{gathered}$$

Distance from top

,$$\begin{gathered} x=h-h\text{'} \\ x=19.6-14.7=4.9m \end{gathered}$$

Hence, the ball goes at the maximum height of $19.6m$, the velocity at which it was thrown is $19.6m{s}^{-1}$,the distance below to its highest point after $3$ sec is $4.9$ m.