Question

Two balls $A$ and $B$ are thrown with speeds $u$ and $\dfrac{u}{2}$ , respectively. Both the balls cover the same horizontal distance before returning to the plane of projection. If the angle of projection of the ball $B$ is $15^\circ$ with the horizontal, then the angle of projection of $A$ is:
A). ${\sin ^{ - 1}}\left( {\dfrac{1}{8}} \right)$
B). $\dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{1}{8}} \right)$
C). $\dfrac{1}{3}{\sin ^{ - 1}}\left( {\dfrac{1}{8}} \right)$
D). $\dfrac{1}{4}{\sin ^{ - 1}}\left( {\dfrac{1}{8}} \right)$

Answer 1

You can start by explaining a projectile. Then move on to explain the horizontal range and write its equation, i.e $R = \dfrac{{{u^2}\sin 2\theta }}{g}$. Then use this equation for both ball $A$ and ball B and then compare these two equations to reach the solution.

Complete step-by-step solution:
Projectile – A projectile is any object or particle that is launched with an initial velocity. The projectile uses this velocity to travel a certain horizontal distance and reach a certain maximum height. The concept of the projectile has been heavily used since old times, it was used in building war tanks, air missiles, and even launching satellites.
Range – It is the horizontal distance that a projectile will cover due to the component of the velocity in the horizontal direction. The horizontal range is not directly affected by the gravitational pull of the Earth.
We know the equation for the range is
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Here, $R =$ The horizontal range
$u =$ The initial speed of the projectile
$\theta =$ The angle that the projectile makes with the horizontal
$g =$ Acceleration due to gravity
Let the angle at which the ball $B$ is thrown be $\theta$
In the problem given to us, we have two projectile balls $A$ and $B$, that are thrown with speeds $u$ and $\dfrac{u}{2}$ , respectively. Both the balls cover the same horizontal distance before returning to the plane of projection.
So, Range of ball $A =$ Range of ball $B$
$\Rightarrow \dfrac{{{{\left( {\dfrac{u}{2}} \right)}^2}\sin 2\left( {15^\circ } \right)}}{g} = \dfrac{{{u^2}\sin 2\theta }}{g}$
$\Rightarrow \dfrac{{{{\left( {\dfrac{u}{2}} \right)}^2}\sin 30}}{g} = \dfrac{{{u^2}\sin 2\theta }}{g}$
$\Rightarrow \dfrac{{{u^2}}}{{8g}} = \dfrac{{{u^2}\sin 2\theta }}{g}$
$\Rightarrow \sin 2\theta = \dfrac{1}{8}$
$\Rightarrow 2\theta = {\sin ^{ - 1}}\left( {\dfrac{1}{8}} \right)$
$\Rightarrow \theta = \dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{1}{8}} \right)$
Hence, option B is the correct option.

Note: In the solution under the horizontal range section, it was mentioned that the gravitational pull of earth does not influence the horizontal range of the projectile directly. But focus on the wording, it does not directly influence it. It influences the maximum height which in turn will affect the horizontal range.