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 proportional to?

Answer 1

Projectile motion will be observed for the given particle because we are projecting it under the influence of gravity. We will use the equations of time taken and range for the projectile motion of the given particle to find the relation for the product of times taken to reach point P in two different ways.

Complete step by step answer:
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 point P in the first case is ${t_1}$.
The time taken to reach 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.
${t_1} = \dfrac{{2u\sin {\theta _1}}}{g}$
The expression for the time taken to reach point P in the second case is:
${t_2} = \dfrac{{2u\sin {\theta _2}}}{g}$……(2)
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:
$r = \dfrac{{{u^2}\sin 2{\theta _1}}}{g}$
For the same value of range in both cases, we can write:

{\theta _1} + {\theta _2} = 90^\circ \\\ {\theta _2} = 90^\circ - {\theta _1} \end{array}$$ On substituting $$90^\circ - {\theta _1}$$ for $${\theta _2}$$ in equation (2), we get: $$\begin{array}{l} {t_2} = \dfrac{{2u\sin \left( {90^\circ - {\theta _1}} \right)}}{g}\\\ = \dfrac{{2u\cos {\theta _1}}}{g} \end{array}$$ The product of time taken to reach point P in both the cases is given by $${t_1}{t_2}$$. We will substitute $$\dfrac{{2u\sin {\theta _1}}}{g}$$ for $${t_1}$$ and $$\dfrac{{2u\cos {\theta _1}}}{g}$$ for $${t_2}$$ in the relation for the product of time taken. $$\begin{array}{l} {t_1}{t_2} = \left( {\dfrac{{2u\sin {\theta _1}}}{g}} \right)\left( {\dfrac{{2u\cos {\theta _1}}}{g}} \right)\\\ = \dfrac{2}{g}\left( {\dfrac{{{u^2}\sin 2{\theta _1}}}{g}} \right) \end{array}$$ Substitute r for $$\dfrac{{{u^2}\sin 2{\theta _1}}}{g}$$ in the above expression. $$\begin{array}{l} {t_1}{t_2} = \dfrac{2}{g}r\\\ {t_1}{t_2} \propto r \end{array}$$ Therefore, the product of the times taken to reach the point P in the two possible ways is directly proportional to the distance r from the point of projection on the same horizontal level. **Note:** We can remember the various equations of motion for a similar type of problem. Do not confuse the acceleration due to gravity (g) while establishing the relationship between the product of times taken and distance r because g is a constant.