Question

The time of flight of a projectile is $10s$ . Range is $500m$ . The maximum height attained by it will be:
(A) $125m$
(B) $50m$
(C) $100m$
(D) $150m$

Answer 1

We start by analyzing the information we have. We then use the formula for time of flight to find the value of the sine component of the initial velocity. We apply this value to the formula of maximum height and get the final answer.

Formulas used: The formula for finding the time of flight of a projectile will be, $T = \dfrac{{2u\sin \theta }}{g}$
Where $g$ is the acceleration due to gravity, its value is, $g = 10m/{s^2}$ (we round off for easier calculation)
The formula for finding the maximum height of a projectile will be, $h = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$

Complete step by step solution:
The following information is given to us,
The time of flight of the projectile is, $T = 10s$
The range of the projectile is, $R = 500m$
Using the value of time of flight, we now find the value necessary in the formula of maximum height reached by the projectile, which is the sine component of initial velocity,
We substitute the values in, $T = \dfrac{{2u\sin \theta }}{g}$ and get
$10 = \dfrac{{2 \times u\sin \theta }}{{10}}$
We separate the value of the sine component of initial velocity to get,
$u\sin \theta = \dfrac{{10 \times 10}}{2} = 50m/s$
Now that we have the value of $u\sin \theta$ we can substitute the value to find the maximum height reached by the projectile by using,
$h = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
$h = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} = \dfrac{{50 \times 50}}{{2 \times 10}} = 125m$
We found the maximum height to be $125m$
In conclusion, the maximum height reached by the projectile will be $125m$ and the right Answer is option (a) $125m$

Note:
Projectile is an object thrown into space whose movement is free under the influence of gravity and air resistance. The easiest example one can think of would be a cricketer throwing a ball to another. Projectile motion is often viewed in sports.