Question

A ball thrown up vertically returns to the thrower after 12 seconds. Find
(i) velocity with which it was thrown up.

Answer 1

Step 1: Given data:

The time for the ball to return is 12 seconds.

Therefore,

⸫ Time taken to reach the maximum height = $\frac{12}{2}$

= 6 seconds

Step 2: Finding the Initial Velocity

Let Initial velocity = u

At the highest point of motion, velocity becomes 0

Therefore

Final velocity ,$v=0$ $\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

And

Acceleration = $g=-10m{s}^{-2}$ (Acceleration is negative because the ball is thrown upwards)

Time = 6 seconds

We know v, a and t

Finding u by 1^st equation of motion

$v=u+at$

$0=u+(-10)\left(6\right)$

$0=u–60$

$u=60m{s}^{-1}$

Thus, the initial velocity is $60m{s}^{-1}$.