Question

A ball thrown vertically upwards falls back on the ground after $6\mathrm{s}$, Assuming that the equation of motion is of the form $\mathrm{s}=\mathrm{ut}–4.9{\mathrm{t}}^2$, where$\text{'}\mathrm{s}\text{'}$ is in meter and $\text{'}\mathrm{t}\text{'}$ is in second, find the velocity at $\mathrm{t}=0$

• A. $0\mathrm{m}/\mathrm{s}$
• B. $1\mathrm{m}/\mathrm{s}$
• C. $29.4\mathrm{m}/\mathrm{s}$
• D. $\mathrm{None}\mathrm{of}\mathrm{these}$

Answer 1

The correct option is C

$29.4\mathrm{m}/\mathrm{s}$

Step 1: Given data

At the highest point of a ball, the velocity of a ball will be equal to zero. Hence we have,

Initial velocity of a ball$=\mathrm{u}$

Final velocity of a ball, $\mathrm{v}=0$

Time taken by ball to return to the ground surface$=6\mathrm{s}$

Hence, time taken to reach the highest position, $\mathrm{t}=3\mathrm{s}$

Given equation of motion is: $\mathrm{s}=\mathrm{ut}-4.9{\mathrm{t}}^2$

Displacement$=\mathrm{s}$

Step 2: Calculating the velocity at $\mathrm{t}=0$

We know that the velocity is equal to the rate of change of displacement. Hence,

$$\begin{gathered} \mathrm{v}=\frac{d\mathrm{s}}{d\mathrm{t}}=\frac{d\left(\mathrm{ut}-4.9{\mathrm{t}}^2\right)}{d\mathrm{t}} \\ \Rightarrow \mathrm{v}=\mathrm{u}-9.8\mathrm{t} \\ \Rightarrow 0=\mathrm{u}-9.8\mathrm{t} \\ \Rightarrow \mathrm{u}=9.8\mathrm{t}=9.8\times 3=29.4{\mathrm{ms}}^{-1} \end{gathered}$$

Therefore initial velocity is equal to $29.4{\mathrm{ms}}^{-1}$

Hence, option(C) is correct answer.