91Ó°ÊÓ

It is given that,

Time taken by the ball to travel from A to$X_1$ and B to$X_2$ is 1s.

Time taken by the ball to travel from$X_1$ to$X_2$ is 4 s.

The horizontal distance between$X_1$ and$X_2$ is 50 m.

We are given the situation that the ball is hit and caught at the same height which is 1meters above ground level. The ball travels from above the wall 1 sec after the hit. The ball is above the wall for 4 sec. We are given that the horizontal distance $X_1$between $X_2$and is 50 m.

As the horizontal velocity of the ball is constant, horizontal acceleration is zero.Kinematic equations can be used to find the horizontal distance between X(initial) and X(final) as well as the initial velocity and angle of projection of the ball.

The first kinematic equation is,

$v=v_0+at$ …(¾±)

The second kinematic equation is,

$s=v_{0}t+\frac{1}{2}a{t}^2$ …(¾±¾±)

The magnitude of vector$=\sqrt{a^2+b^2}$ …(¾±¾±¾±)

Angle made by velocity vector with the x-axis,

role="math" localid="1661147007012" $\mathrm{θ}={\tan }^{-1}\left(\frac{V_y}{V_x}\right)$ …(¾±±¹)

Let’s start from the horizontal distance. As the velocity in the horizontal direction is constant, we can use the 2^nd kinematic equation here for horizontal distance which is above the wall, and we can conclude the initial velocity component in that direction.

$$\begin{gathered} s=v_{0}t+\frac{1}{2}g{t}^2 \\ {X}_2-X_1=v_{0}t×t \\ 50=v_{0}t×4 \\ {v}_{ox}=12.5m/s \end{gathered}$$

The time required to travel from A to $X_1$is 1 and $X_2$ to B is 1 s. The time taken by the ball to travel from $X_1$to $X_2$is 4 s. Therefore, the total time of flight is 6 s. If we solve the same equation for X (initial) to X (final) by using the above value of the initial horizontal velocity component, we can find the horizontal distance as,

$$\begin{gathered} X\left(final\right)-X\left(initial\right)=(12.5m/s)×6s+\frac{1}{2}\left(0\right)×{\left(6s\right)}^2 \\ X\left(final\right)-X\left(initial\right)=75m \end{gathered}$$

Therefore, horizontal distance traveled = 75 meters.

Now, we applythesame kinematic equation for the vertical motion of the ball for findingtheinitial vertical velocity component. But for that, we have to first findthevertical velocity oftheball above the wall.

$$\begin{gathered} s=v_{0}t+\frac{1}{2}g{t}^2 \\ {X}_1-X_2=v_{1y}t-\frac{1}{2}(9.8m/s^2)t^{2}forverticalmotion \\ 0=v_{1y}(4s)-\frac{1}{2}(9.8m/s^2)\left(16s^2\right) \\ {v}_{1y}=19.6m/s \end{gathered}$$

This is the vertical velocity of the ball above the wall, but we need the initial vertical velocity component. If we use the 1^st kinematic equation and as we have velocity above the wall, so we can use it as the final velocity, and our desired velocity is initial. Also, we have the time of 1 sec.

$$\begin{gathered} v=v_{0y}+at \\ 19.6m/s=v_{0y}-\left(9.8mi{s}^2\right)(1s) \\ {v}_{0y}=29.4m/s \end{gathered}$$

So, the vertical component of velocity is 29.4 m/s.

Now, the magnitude of initial velocity is

$$\begin{gathered} v_0=\sqrt{{\left(12.5m/s\right)}^2+{\left(29.4m/s\right)}^2} \\ =31.9m/s \end{gathered}$$

So, the magnitude of initial velocity is 3.19 m/s.

Now, the angle of the initial velocity vector with the x-axis is,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{29.4m/s}{12.5m/s}\right) \\ =67° \end{gathered}$$

Thus, the angle made by the initial velocity vector with the x-axis is$67°$.

Now, the height of the wall above the point of impact.

$\begin{array}{l}s=v_{0}t+\frac{1}{2}g{t}^2\\ =29.4m/s\left(1s\right)-\frac{1}{2}(9.8m/s^2){(1s)}^2\\ =24.5m\end{array}$

localid="1661148502998" $$\begin{gathered} Totalheight=heightabovetheimpact+heightofimpactfromground \\ =(24.5m+1m) \\ =25.5m \end{gathered}$$.

Thus, Height of the wall is 25.5 m.