Question

A ball is thrown vertically upwards with a velocity of $49m{s}^{-1}$. Calculate the maximum height to which it rises.

Answer 1

Step 1: Given Data

The initial velocity of the ball $u=49m{s}^{-1}$

The final velocity of the ball $v=0$

Let the acceleration due to gravity be $g=9.8m{s}^{-2}$

Let the maximum height be $h$.

Step 2: Formula used

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

Step 3: Calculate the Height

According to Newton's third equation of motion

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

Acceleration is given as negative since the ball moves upwards against gravity.

$0={49}^2-2\times 9.8\times h$

$\Rightarrow {49}^2=2\times 9.8\times h$

$\Rightarrow h=\frac{{49}^2}{2\times 9.8}$

$\Rightarrow h=122.5m$

Hence, the maximum height to which the ball rises is $122.5m$.