You are to throw a ball with a speed of $\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$at a target that is height
$\mathit{h}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$above the level at which you release the ball (Fig. 4-58). You want the ball’s velocity to be horizontal at the instant it reaches the target. (a) At what angle above the horizontal must you throw the ball? (b) What is the horizontal distance from the release point to the target? (c) What is the speed of the ball just as it reaches the target?

It is given that,

$v_0=12.0\mathrm{m}/\mathrm{s}$

$h=5.00\mathrm{m}$

$a=9.8\mathrm{m}/{\mathrm{s}}^2$

This problem involves the projectile motion of an object. Also it is based on the resolution of components of vector. The resolution of a vector is the splitting of a single vector into two or more vectors in different directions. To initiate, the angle of projection can be found by using trigonometric functions and given velocity. And further, resolve the velocity into x and y components. Using the given height of the target, the time can be found.

Formulae:

The velocities in kinematic equations are given as,

$v_f=v_0+at$ (i)

localid="1657014436706" $v_f^2=v_0^2+2ad$ (ii)

$d=v_{0}t+\left(\frac{1}{2}\right)a{t}^2$ (iii)

where, d is total displacements,$v_{0}and{v}_f$ are the initial and final velocities, t is time and a is an acceleration.

Now, resolve velocity in x and y components,

$\begin{array}{ll}{v}_{0x}=v_0& \cos \theta \\ v_{0y}=v_0& \sin \theta \end{array}$

The final velocity at the top of the projectile motion is 0. Thus by substituting the given values in equation (ii),

$0={\left(V_{0y}\right)}^2+2ad$

localid="1657010768330" $0={(v_0\sin \theta )}^2+2(-9.8)(5.00)$

$0={\left(12.0\sin \theta \right)}^2-98$

$144{\left(\sin \theta \right)}^2=98$

${(\sin \theta )}^2=\frac{98}{144}$

localid="1657014874801" $=0.6805$

$\sin \theta =\sqrt{0.6805}$

$=0.8249$

localid="1657015988392" $\begin{array}{c}\theta =\left(0.8249\right)\\ \begin{array}{c}=55.{58}^0\end{array}\end{array}$

Angle of projection of the ball islocalid="1657011453849" $55.{58}^0$

Using equation (i),

$0=v_{0y}+at$

$\mathrm{𝟘}=12\sin (55.58)+(-98)t$

$9.8t=9.8999$

localid="1657013526529" $\begin{array}{l}t=\frac{9.8999}{9.8}\\ =1.01\mathrm{s}\end{array}$

Using this time of flight in the third equation we can find horizontal distance, here, the acceleration in x direction is zero. Thus $a_x=0$

$d=v_{0x}t$

$d=\left(12\cos \left(55.58\right)\right)\left(1.01\right)$

$=6.85\mathrm{m}$

Therefore, the horizontal distance is$=6.85\mathrm{m}$.

As there is a_x = 0, the horizontal velocity component remains the same value and vertical velocity component becomes 0 at the target.

Therefore, the velocity at the target is initial horizontal velocity component that is

$v_{0x}=12\cos \left(55.58\right)$

$=6.78\mathrm{m}/\mathrm{s}$

Therefore, the speed of the ball at the target is $=6.78\mathrm{m}/\mathrm{s}$