Question

A ball thrown up is caught back by the thrower after $6s$. Calculate (i) the velocity with which the ball was through up, (ii) the maximum height attained by the ball, and (iii) the distance covered by the ball after $2s$. $(Take,g=9.8m{s}^{-2})$

Answer 1

Step 1: Given data:

Time to reach maximum height, $t=\frac{6}{2}=3s$

Gravity, $g=9.8m{s}^{-2}$

Final velocity, $v=0$

Step 2: Finding the velocity with which it was thrown up

$v=u+gt$

Substituting the values into the formula

$0=u+9.8\times 3$

$u=29.4m{s}^{-1}$

Therefore, the velocity with which the ball was thrown up is $29.4m{s}^{-1}$

Step 3: finding the maximum height reached is given by

$2gs=v^2-u^2$

Substituting the values into the formula

$-(2\times 9.8\times s)=0-29.4\times 29.4$

$s=44.1m$

Therefore, the maximum height reaches by it is $44.1m$.

Step 4: Finding the Distance covered

Distance covered by the ball after $2s$ is given by $s=ut-\frac{1}{2}g{t}^2$

$=29.4\times 2-\frac{1}{2}\times 9.8\times 2\times 2$

$s=39.2m$

The distance covered by the ball after $2s$ is $39.2m$.