Question

A ball $\mathrm{P}$ is dropped vertically and another ball $\mathrm{Q}$ is thrown horizontally with the same velocities from the same height and at the same time. If air resistance is neglected, then

• A. Ball $\mathrm{P}$ reaches the ground first
• B. Ball $\mathrm{Q}$ reaches the ground first
• C. Both reach the ground at the same time
• D. The respective masses of the two balls will decide the time

Answer 1

The correct option is C

Both reach the ground at the same time

Step 1: Given data

It is given in the question that the velocities of both the balls is same and both the balls are at same height.

Both the balls are thrown at same time, that is:

Ball $\text{'}\mathrm{P}\text{'}$is thrown vertically downward and ball $\text{'}\mathrm{Q}\text{'}$is thrown horizontally with same velocities.

Step 2: Calculating the time to reach the ground.

We know that according to the equation of motion:

$\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2$

$\text{'}\mathrm{S}\text{'}$ is the height of ball from the ground

$\text{'}\mathrm{u}\text{'}$is the initial velocity of a ball which is equal to zero.

$\text{'}\mathrm{g}\text{'}$is the acceleration due to gravity.

$\text{'}\mathrm{t}\text{'}$is the time taken to reach the ground

Hence using the above equation of motion, we get

$$\begin{gathered} \mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2 \\ \Rightarrow \mathrm{S}=0+\frac{1}{2}{\mathrm{gt}}^2 \\ \Rightarrow {\mathrm{t}}^2=\frac{2\mathrm{S}}{\mathrm{g}} \\ \Rightarrow \mathrm{t}=\sqrt{\frac{2\mathrm{S}}{\mathrm{g}}} \end{gathered}$$

As we can see that, the time only depends on height of a balls which is same in our case for both the balls. Therefore both the balls will reach the ground at same time.

Hence, option(C) is the correct answer.