Question

It is possible to project a particle with a given speed in two possible ways such that it hits a point P at a distance r from the point of projection on the same horizontal level. The product of the times taken to reach this point in the two possible ways is equal to?

• A. $\frac{4r}{g}$
• B. $\frac{2r}{g}$
• C. $\frac{3r}{g}$
• D. $\frac{r}{g}$

Answer 1

The correct option is B

$\frac{2r}{g}$

Step 1: Given data

Given that the velocity of projection for both the cases is the same.

Step 2: Projectile motion

1. Because we are projecting the particle under the influence of gravity, the projectile motion will be detected.
2. In two distinct techniques, we will use the time taken and range equations for the projectile motion of the supplied particle to get the relation for the product of times taken to reach point $P$.

Step 3: Formula used

$r=\frac{u^2\sin 2{\theta }_1}{g}[wherer=range,u=velocityofprojection,g=gravitationalcons\tan t,\theta =anglebetweenpointofparticleandpointofproject$

Step 4: Calculating time taken

Let us assume,

The angle of projection in the first case is ${\theta }_1$

The angle of projection in the second case is ${\theta }_2$

The time taken to reach a point $P$ in the first case is $t_1$

The time taken to reach a point $P$ in the second case is $t_2$

It is given that the velocity of projection for both the cases is the same as can be assumed as $u$

Let us write the expression for the time taken to reach point $P$ in the first case.
$t_1=\frac{2u\sin {\theta }_1}{g}$

The expression for the time taken to reach a point $P$ in the second case is:
$t_2=\frac{2u\sin {\theta }_2}{g}......\left(2\right)$

Step 5: Calculating the value of the range

It is also given that the horizontal distance between point $P$ and point of projection is r, and we know that this distance is called range.

We can write the expression for range for the first case as below:
$r=\frac{u^2\sin 2{\theta }_1}{g}$

For the same value of range in both cases, we can write

$$\begin{gathered} {\theta }_1+{\theta }_2={90}^o \\ {\theta }_2={90}^o-{\theta }_1 \end{gathered}$$

Step6: Concluding

On substituting ${90}^o-{\theta }_{1}for{\theta }_2$in equation $\left(2\right)$, we get

$$\begin{gathered} t_2=\frac{2u\sin ({90}^o-{\theta }_1)}{g} \\ {t}_2=\frac{2u\cos {\theta }_1}{g} \end{gathered}$$

The product of time taken to reach point $P$ in both the cases is given by$t_1{t}_2$.

We will substitute $\frac{2u\sin {\theta }_1}{g}for{t}_{1}and\frac{2u\cos {\theta }_1}{g}for{t}_2$ in the relation for the product of time taken.
$t_1{t}_2=\left(\frac{2u\sin {\theta }_1}{g}\right)\left(\frac{2u\cos {\theta }_1}{g}\right)=\frac{2}{g}\left(\frac{u^2\sin 2{\theta }_1}{g}\right)$

Substitute r for $\frac{u^2\sin 2{\theta }_1}{g}$ in the above expression.
$t_1{t}_2=\frac{2}{g}r$

Hence, option B is the correct answer.