Question

A ball is thrown up with a speed of $9.8m{s}^{-1}$. What is the time taken by the ball to reach back to the starting point from the highest point reached?

Answer 1

Step 1: Given:

Initial velocity, $u=0m/s$

At the highest point the ball comes to rest for a while and after that starts falling under the effect of gravity with the speed at which it is thrown that is $9.8m/s$.

Final velocity,$v=9.8m/s$

Acceleration due to gravity, $g=9.8m/s^2$

Step 2: Finding the time taken by the ball to reach back to the starting point from the highest point reached:

By the 1st equation of motion is

$$\begin{gathered} v=u+at \\ a=g=9.8m/s^2 \\ v=u+gt \end{gathered}$$

Putting the values in the above equation.

$$\begin{gathered} (9.8m/s)=(0m/s)+(9.8m/s^2)\times t \\ t=\frac{9.8}{9.8} \\ t=1s \end{gathered}$$

Hence, the time taken by the ball to reach back to the starting point from the highest point is $1second$.