Two seconds after being projected from ground level, a projectile is displaced$\mathbf{40}\mathbf{}\mathbf{m}$horizontally and $\mathbf{53}\mathbf{}\mathbf{m}$vertically above its launch point. What are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c)At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from the launch point?

It is given that, the time is $2\mathrm{s}$ after projection. Projectile attends $4\mathrm{m}$ and $54\mathrm{m}$horizontal and vertical distance respectively.

The ball travels some vertical and horizontal distance after $\mathbf{2}\mathbf{s}$of projection. At the time of projection, the ball is projected with some angle, so resolving that initial velocity with the given angle along X and Y axis, then the terms in form of

$\mathbf{\theta }$ can be obtained. For horizontal distance, use the 2nd kinematic equation as horizontal journey has zero acceleration. So, the horizontal component of initial velocity can be found, and by using the same equation, the vertical component of initial velocity can also be obtained.

With the help of 1stand 2ndkinematic equation, find the horizontal distance at the maximum height.

Formula:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\mathrm{o}}+\mathrm{at} \\ \mathrm{s}={\mathrm{v}}_{\mathrm{o}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \end{gathered}$$ (i)

Let’s take the origin as the point directly below to the impact point. To find the height of the ball at a given horizontal distance, its initial velocity and time to reach that given horizontal distance need to find.

Using equation (ii) the horizontal distance is given by,

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_{\mathrm{o}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ 40=v_o\cos \theta \times 2=0 \\ {\mathrm{v}}_{\mathrm{o}}\mathrm{cos\theta }=20\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (ii) the vertical distance is given by,

role="math" localid="1654159530981" $$\begin{gathered} \mathrm{y}={\mathrm{v}}_{\mathrm{o}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ 53=v_o\mathrm{sin\theta }\times 2-\left(\frac{1}{2}\times 9.8\times 4\right) \\ {v}_o\mathrm{sin\theta }=36.3\mathrm{m}/\mathrm{s} \end{gathered}$$

But, the time for horizontal distance and no acceleration in horizontal direction are needed.

So,

$$\begin{gathered} x={\mathrm{v}}_{\mathrm{o}}\mathrm{cos\theta }\times \mathrm{t} \\ \mathrm{t}=\frac{\mathrm{x}}{{\mathrm{v}}_{\mathrm{o}}\mathrm{cos\theta }} \\ \mathrm{t}=4.258\mathrm{s} \end{gathered}$$

Now, vertical component of initial velocity had been obtained, and the velocity of projectile is zero at maximum height. So, find the time to reach the maximum height by using 1stkinematic equation,

$$\begin{gathered} v_{\mathrm{f}}={\mathrm{v}}_{\mathrm{o}}+\mathrm{at} \\ \mathrm{o}=36.3-9.8\mathrm{t} \\ \mathrm{t}=3.70\mathrm{s} \end{gathered}$$

By using this time, find the position of projectile at horizontal distance.

By using 2ndkinematic equation,

$$\begin{gathered} \mathrm{s}={\mathrm{v}}_{\mathrm{o}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{s}=20\times 3.70 \end{gathered}$$

Thus, $\mathrm{s}=74\mathrm{m}$