Question

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

Answer 1

Step 1: Given data:

Initial velocity $\left(u\right)=9.8m{s}^{-1}$

Final velocity $\left(v\right)=0m{s}^{-1}$

Acceleration due to gravity $\left(g\right)=-9.8m{s}^{-2}$

(Note, the ball is moving against gravity, thus, the value of acceleration due to gravity is taken as a negative value.)

Step 2: Finding the time taken by the ball to reach the highest point:

By the first equation of motion is

$v=u+at$

$a=-g=-9.8m{s}^{-2}$

Putting the given values in the equation:

$$\begin{gathered} 0=9.8-9.8\times t \\ -9.8t=-9.8 \\ t=1s \end{gathered}$$

Hence, the time taken by the ball to reach the highest point is $1sec$.