Question

From a point on the ground at a distance of 2m from the foot of a vertical wall, a ball is thrown at an angle of $45^\circ$ which just clears the top of the wall and afterward strikes the ground at a distance of 4m on the other side. The height of the wall is
A) $3/2$
B) $2/3$
C) $3/4$
D) $4/3$

Answer 1

Projectile motion is one specific form of motion where the object moves in a bilaterally symmetrical, parabolic path. For the ball to clear the wall, it must be projected such that its maximum height attained is just equal to the height of the wall.
Formula used: In this solution, we will use the following formula:
Vertical displacement of a projectile: $y = xtan\theta - \dfrac{{g{x^2}}}{{2{u^2}co{s^2}\theta }}$ where $x$ is the horizontal displacement, $g$ is the gravitational acceleration , $\theta$ is the angle of projection, and $u$ is the velocity of projection
Range of a projectile: $R = \dfrac{{{u^2}\sin 2\theta }}{g}$

Complete step by step answer
We’ve been told that a ball is thrown at an angle of $45^\circ$ at a wall which is 2 metres away. The ball is thrown such that the ball clears a wall of height 2m and strikes the ground 4 metres away from its initial position. This means that the wall is at the centre of the path of the projectile trajectory of the ball and that the range of the projectile is $2 + 4 = 6$ metres. So, using the formula for range of a projectile, we get:
$6 = \dfrac{{{u^2}\sin 90^\circ }}{{9.8}}$
Which gives us
${u^2} = 58.8$
Taking the square root on both sides, we get
$u = 7.6\,m/s$
The vertical displacement of a projectile is given as
$y = xtan\theta - \dfrac{{g{x^2}}}{{2{u^2}co{s^2}\theta }}$
Now the wall is located 2 metres away from the starting position so substituting $x = 2$ in the above equation, we get
$y = 2\tan 45^\circ - \dfrac{4}{{2{{\cos }^2}45^\circ }}\left( {\dfrac{1}{6}} \right)$
Solving the above equation, we get
$y = \dfrac{4}{3}m$
This will be the height of the wall for the ball to pass above it hence the correct choice is option (D).

Note
Since the wall is not at the midpoint of the projectile motion horizontal position, we must use the formula of vertical displacement of projectile motion. We must realize that the height of the wall indirectly corresponds to the vertical displacement that the projectile must have to just clear the top of the wall.