Question

A particle is projected under gravity at an angle of projection $45^\circ$ with horizontal. Its horizontal range is $36\;{\text{m}}$. Find the maximum height attained by the particle.
A) $8\;{\text{m}}$
B) $9\;{\text{m}}$
C) $10\;{\text{m}}$
D) $20\;{\text{m}}$

Answer 1

In the given question, the concept of the maximum height attained by the particle in the projectile motion and the horizontal range of the projectile motion will be used. Use the standard formula of the f maximum height and range of the projectile motion to obtain the maximum height of the particle.

Complete step by step solution:
A projectile is an object that is given an initial velocity and is acted on by gravity. The maximum height of the object is the highest vertical position along its trajectory. The maximum height of the projectile depends on the initial velocity $v$, the launch angle $\theta$, and the acceleration due to gravity. The unit of maximum height is meter.
Now, we write the formula for the maximum height for the projectile motion as,
${H_{\max }} = \dfrac{{{V_0}^2si{n^2}\theta }}{{2g}}......(i)$
Where, $H$ is the maximum height, ${V_0}$ is the initial velocity, $g$ is the acceleration due to gravity, and $\theta$ is the angle of the initial velocity from the horizontal plane.
Now, we write the formula for the horizontal range of the projectile motion as,
$R = \dfrac{{{V_0}^2sin2\theta }}{g}......(ii)$

Here, $R$ is the horizontal range.
Now, from the trigonometry formula, we know: $\sin 2\theta = 2\sin \theta \cos \theta$
So, by putting the value of $\sin 2\theta$ in equation (ii) we get,
$\Rightarrow R = \dfrac{{{V_0}^22\sin \theta \cos \theta }}{g}......(iii)$
Now by dividing equation $\;\left( {iii} \right)$ by equation $\left( i \right)$ we get

$$\dfrac{R}{H} = \dfrac{{\dfrac{{{V_0}^22\sin \theta \cos \theta }}{g}}}{{\dfrac{{{V_0}^2{{\sin }^2}\theta }}{{2g}}}} \\\ \Rightarrow \dfrac{R}{H} = 4\cot \theta ......(iv) \\\$$

Now according to question, $\theta = 45^\circ$ (given)
So, by putting the value of $\theta$ in equation (iv) we get:
$\dfrac{{36}}{H} = 4\cot 45^\circ$
(As $\cot {45^ \circ } = 1$)
$\Rightarrow H = \dfrac{{36}}{4}$
On simplification we get,
$\therefore H = 9\;{\text{m}}$
Thus, the maximum height attained by the particle is $9\;{\text{m}}$

Hence, the correct answer is option (B).

Note: In the above solution, alternatively the height can also be divided by range to get the simplified formula. In that case, in terms of $\tan \theta$ we will get the simplified version and in a similar way, we can calculate the angle.