Question

When a particle is projected at some angles to the horizontal, it has a range R and time of flight \[{{t}_{1}}\]. If the same particle is projected with the same speed at some other angle to have the same range its time of flight is \[{{t}_{2}}\], then,
A- \[{{t}_{2}}+{{t}_{1}}=\dfrac{2R}{g}\]
B- \[{{t}_{1}}-{{t}_{2}}=\dfrac{R}{g}\]
C- \[{{t}_{1}}{{t}_{2}}=\dfrac{2R}{g}\]
D- \[{{t}_{1}}{{t}_{2}}=\dfrac{R}{g}\]

Answer 1

Hint: This problem talks about projectile motion. Initially, a particle is projected at some angle to the horizontal and the range is R and time is \[{{t}_{1}}\]. Now the particle is projected at the same speed, different angle and the range is the same R but time is \[{{t}_{2}}\].

Complete step by step answer:For the first case:
Let the angle be \[\alpha \]
The range will be given as \[R=\dfrac{{{u}^{2}}\sin 2\alpha }{g}\]-----(1)
For the second case:
Let the angle be \[\beta \]
The range will be given as \[R=\dfrac{{{u}^{2}}\sin 2\beta }{g}\]----(2)
In eq (1) and (2) LHS are the same, so
\[\dfrac{{{u}^{2}}\sin 2\beta }{g}=\dfrac{{{u}^{2}}\sin 2\alpha }{g}\]
\[\sin 2\beta =\sin 2\alpha \]
That implies that one angle is greater by others by \[90{}^\circ \]
Thus \[\alpha =90{}^\circ -\beta \]
Now for time,
$
{{t}_{1}}=\dfrac{2u\sin (90{}^\circ -\beta )}{g}=\dfrac{2u\cos \beta }{g} \\
\implies {{t}_{2}}=\dfrac{2u\sin \beta }{g} \\
$
Multiplying both the times,
$
{{t}_{1}}{{t}_{2}}=\dfrac{2u\cos \beta }{g}\times \dfrac{2u\sin \beta }{g} \\
\implies {{t}_{1}}{{t}_{2}}=\dfrac{2{{u}^{2}}\sin 2\beta }{{{g}^{2}}} \\
\therefore {{t}_{1}}{{t}_{2}}=\dfrac{2R}{g} \\
$
So, the correct option is (C).

Additional Information: Projectile motion is the motion of an object thrown into the air, with some angle with the horizontal and subject to only the acceleration of gravity. There is no acceleration in the horizontal direction and so the horizontal velocity remains constant, the acceleration remains constant throughout the motion of the object.

Note:
This was a good example of using the equations of motions of the projectile while solving the trigonometric equation. We have considered the fact that the value of sine is positive in the first and the second quadrant.