Question

A ball thrown up vertically returns to the thrower after $6s$. Find

(a) The velocity with which it was thrown up.

Answer 1

Step 1: Given data

Total time of journey, $T=6s$.

Time of ascent $t=\frac{T}{2}=3s$.

The final velocity of the ball during ascent, $v=0m/s$.

Acceleration during the ascent, $a=-g=-10m/s^2$.

Step 2: Formula for calculating initial velocity during ascent

$$\begin{gathered} \mathrm{v}=\mathrm{u}+\mathrm{at} \\ \Rightarrow \mathrm{u}=\mathrm{v}-\mathrm{at} \\ \mathrm{Where}, \\ \mathrm{v}\to \mathrm{Final}\mathrm{velocity} \\ \mathrm{u}\to \mathrm{Initial}\mathrm{velocity} \\ \mathrm{a}\to \mathrm{Acceleration} \\ \mathrm{t}\to \mathrm{Time} \end{gathered}$$

Step 3: Calculating initial velocity during ascent

$$\begin{gathered} u=v-at \\ \Rightarrow u=0-(-10\times 3) \\ \Rightarrow u=30m/s \end{gathered}$$

Thus, the velocity with which it was thrown up is $30m/s$.