You throw a ball toward a wall at speed${}^{\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$and at angle${}^{{\mathit{\theta }}_{\mathbf{0}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{°}}$above the horizontal (Fig. 4-35). The wall is distance${}^{\mathbf{d}\mathbf{=}\mathbf{22}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}}$from the release point of the ball. (a)How far above the release point does the ball hit the wall? What are the (b)horizontal and (d) vertical components of its velocity as it hits the wall? (e)When it hits, has it passed the highest point on its trajectory?

$$\begin{gathered} v_0=25.0m/s \\ \theta =40.0° \end{gathered}$$

Distance of the wall from release point is${}^{d_x=22.0m}$

localid="1654583112888" $Acceleration=GravitationalAcceleration=-9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}$

This problem deals with the kinematic equations that describe the motion of an object at constant acceleration.

With the given values of velocity and angle, thetime required to hit the wall can be found. Using the calculated time, the distance how high will it go in that timewill be calculated. Further, using kinematic equations, the final velocity when ball is about to hit the ball can be found.

Formulae:

The final velocity in kinematic equation can be written as

$v_f^2=v_0^2+2a{d}_y$ (i)

Where, ${}^{d_y}$is the vertical distance which is given by

$d_y=v_{0}t+\frac{1}{2}a{t}^2$ (ii)

The speed is,

$v=\frac{d}{1}$ (iii)

The velocity on horizontal direction is,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{ox}}=25\times \cos 40 \\ =19.15\mathrm{m}/\mathrm{s} \end{gathered}$$

The velocity on vertical direction is,

$$\begin{gathered} {\mathrm{v}}_{0\mathrm{y}}=25\times \sin 40.0 \\ =16.07\mathrm{m}/\mathrm{s} \end{gathered}$$

With equation (iii), the time required to reach the wall is,

$$\begin{gathered} \mathrm{t}=\frac{{\mathrm{d}}_{\mathrm{x}}}{{\mathrm{v}}_{\mathrm{ox}}} \\ =\frac{22.0}{19.15} \\ \mathrm{t}=1.15\mathrm{s} \end{gathered}$$

With this time the equation (ii) can be given as,

$$\begin{gathered} {\mathrm{d}}_{\mathrm{y}}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ =\left(16.07\times 1.15\right)+\left(\frac{1}{2}\times 9.8\times 1.{15}^2\right) \\ =18.48-6.48 \end{gathered}$$

Thus,

${\mathrm{d}}_{\mathrm{y}}=12.0\mathrm{m}$

As there is no acceleration acting in x- direction, the final velocity in the x direction will be the same as that of initial.

$v_{tx}=19.2m/s$

Now using equation (i)

$$\begin{gathered} v_{fy}^2=v_{0y}^2+2a{d}_y \\ {v}_{fy}^2=\sqrt{v_{0y}^2+2d_y} \\ =\sqrt{16.{07}^2-2\times 9.8\times 12} \\ =\sqrt{258.25-235.2} \\ =\sqrt{23.05} \end{gathered}$$

Thus

${\mathrm{v}}_{\mathrm{fy}}=4.8\mathrm{m}/\mathrm{s}$

When the ball reaches the maximum point of trajectory, its final velocity in the y-direction will be zero. But in this case, the final velocity of the ball when it hits the wall is greater than zero. Thus,the ball yet not reached the highest point of the trajectory.

Thus, the trajectory of the ball can be found using the equation for the path.