A stone is projected from the ground with the velocity $50m/s$ at an angle ${30^ \circ }$ . It crosses the wall after$4\sec $. How far beyond the wall the stone will strike the ground.
$\left( a \right)$ 25m
$\left( b \right)$ $25\sqrt 3 m$
$\left( c \right)$ $50m$
$\left( d \right)$ $\dfrac{{25}}{{\sqrt 3 }}m$

Hint: The time of flight of a projectile motion is that the time from once the article is projected to the time it reaches the surface. T depends on the initial velocity magnitude and also the angle of the projectile. The range of motion is fastened by the condition.

Formula used:
Time of flight,
$ \Rightarrow T = \dfrac{{2u\sin \theta }}{g}$
Where $T$ will be the time of flight, $u$ will be the initial velocity, and $g$ will be the acceleration due to gravity.
Range of projectile,
$ \Rightarrow R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Where $R$ will be the range of the projectile and $u$will be the initial velocity and $g$ will be the acceleration due to gravity.

Complete step by step solution:
Since we know the formula for finding the time of flight
So we will put the values in the below formula,
$ \Rightarrow T = \dfrac{{2u\sin \theta }}{g}$
Now on putting the values, we get
$ \Rightarrow \dfrac{{2 \times 50 \times \sin {{30}^0}}}{{10}}$
Further solving the above,
$ \Rightarrow \dfrac{{100 \times \left( {1/2} \right)}}{{10}}$
$ \Rightarrow 5s$
Now we will find the range of the projectile
So for this also we have already known the formula. Now we will put the values in the formula
$ \Rightarrow R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Putting the values, we get
$ \Rightarrow \dfrac{{2500 \times \left( {\sqrt 3 /2} \right)}}{{10}}$
Further solving we will get the value for the range of trajectory,
$ \Rightarrow 216.5m$
Now we have time$t = 4\sec $,
So by equating the equation
\[ \Rightarrow \dfrac{t}{T} = \dfrac{X}{R}\]
Now we will fill the known values, we get
$ \Rightarrow \dfrac{4}{5} = \dfrac{X}{{216.5}}$
On solving the above equation
$ \Rightarrow X = \dfrac{4}{5} \times 216.5$
$ \Rightarrow 173.2m$
Now we will calculate the distance the stone strikes the ground.
Therefore,
$ \Rightarrow R - X = 216.5 - 173.2$
$ \Rightarrow 43.3m = 25\sqrt 3 m$
So above is the distance required to hit the ground.

Note: Projectile motion may be a type of motion wherever an object moves in a very bilaterally symmetrical, parabolic path. The path that the thing follows is termed its trajectory. Projectile motion solely happens once there's one force applied at the start of the trajectory, once that the sole interference is from gravity.