Question

A projectile initially has the same horizontal velocity as it would acquire, if it had moved from rest with uniform acceleration of $3m{{s}^{-2}}$ for 0.5 minutes. If the maximum height reached by it is 80m, then the angle of projection is ($g=10m{{s}^{-2}}$)
A. ${{\tan }^{-1}}\left( 3 \right)$
B. ${{\tan }^{-1}}\left( \dfrac{3}{2} \right)$
C. ${{\tan }^{-1}}\left( \dfrac{4}{9} \right)$
D. ${{\sin }^{-1}}\left( \dfrac{4}{9} \right)$

Answer 1

Hint:Calculate the initial horizontal velocity of the projectile with the help of the given information. Then use the relation between the initial horizontal velocity of the projectile and the angle of projection to find the initial velocity. Finally, use the formula for the maximum height and substitute the value of the initial velocity to calculate the angle of projection.

Formula used:
$H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$
$v=u+at$

Complete step by step answer:
Let us assume that the projectile is projected from the ground level, i.e. it is projected from a height of 0 m.

Let the angle of projection be $\theta $, the initial velocity be u.

The maximum height attained by the projectile according to the assumption is given as
$H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$ ….. (i), where $g$ is the acceleration due to gravity.

Let us calculate the initial velocity of the projectile.
It is given that the initial horizontal velocity of the projectile is the same as the velocity acquired by it when it is uniformly accelerated from rest with an acceleration of $3m{{s}^{-2}}$ for 0.5 minutes.

Let the initial horizontal velocity of the projectile be ${{u}_{x}}$.
For a constant acceleration, $v=u+at$, where u is the initial velocity of the particle, a is its
acceleration and v is its velocity after accelerating it for time t.

Now, substitute $v={{u}_{x}}$, $u=0$, $a=3m{{s}^{-2}}$ and $t=0.5\min =30s$.
$\Rightarrow {{u}_{x}}=0+(3)(30)=90m{{s}^{-1}}$.

This means that the initial horizontal speed of the projectile is $90m{{s}^{-1}}$.
In the projectile motion, ${{u}_{x}}=u\cos \theta $.
$\Rightarrow u=\dfrac{{{u}_{x}}}{\cos \theta }$.
Substitute this value of u in (i).
$\Rightarrow H=\dfrac{{{\left( \dfrac{{{u}_{x}}}{\cos \theta } \right)}^{2}}{{\sin }^{2}}\theta }{2g}$

$\Rightarrow H=\dfrac{u_{x}^{2}\left( \dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }
\right)}{2g}=\dfrac{u_{x}^{2}{{\tan }^{2}}\theta }{2g}$

Now, substitute the known values.
$\Rightarrow 80=\dfrac{{{\left( 90 \right)}^{2}}{{\tan }^{2}}\theta }{2(10)}$
$\Rightarrow {{\tan }^{2}}\theta =\dfrac{80\times 20}{90\times 90}=\dfrac{16}{81}$
$\Rightarrow \tan \theta =\pm \sqrt{\dfrac{16}{81}}=\pm \dfrac{4}{9}$

We discard the negative value because the angle of the projectile is always positive.
$\Rightarrow \tan \theta =\dfrac{4}{9}$
$\Rightarrow \theta ={{\tan }^{-1}}\dfrac{4}{9}$.

Hence, the correct option is C.

Note:If you do not know the formula for the maximum height attained by a projectile, then you can easily derive it by using the suitable kinematic equations.

You may use the kinematic equation $2as={{v}^{2}}-{{u}^{2}}$, where displacement of the projectile.

You can apply this equation to the vertical motion and use the fact that the velocity of the projectile at the maximum height is zero.