Question

A ball is thrown up with a speed of $9.8m{s}^{-1}$. What is the maximum height reached?

Answer 1

Step 1: Given data:

Initial velocity, $u=9.8m/s$

Final velocity, $v=0m/s$ (As the ball comes to rest at the highest point)

Acceleration due to gravity, $g=-9.8m/s^2$ (As the ball is thrown up, the direction of acceleration due to gravity is opposite to the direction of motion of the ball, thus, the value of acceleration is taken as negative).

Step 2: Finding the maximum height:

By the third equation of motion,

$v^2-u^2=2as$

Here, $a=g$and $s=h$

So, $v^2-u^2=2gh$

Putting the values in the above equation.

$$\begin{gathered} {\left(0m/s\right)}^2-{\left(9.8m/s\right)}^2=2\times \left(-9.8m/s^2\right)\times h \\ -{\left(9.8\right)}^2=-2\times 9.8\times h \\ h=\frac{-{\left(9.8\right)}^2}{-19.6} \\ h=4.9m \end{gathered}$$

Hence, the maximum height reached is $4.9m$.