In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 slater at height h=20.0 mabove the release level. The ball’s path just before landing is angled at$\mathit{\theta }\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$with the roof. (a) Find the horizontal distanceit travels. (See the hint to Problem 39) What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball’s initial velocity?

It is given that, the height of the roof is 20.0 m and angle of landing is $60°$and time is 4.0 s to travel the total distance.

As the ball is hit on the roof with a given angle of projection, it is a time-reversed problem with the ball thrown from the roof towards left. The height of the roof is also given. So, kinematic equations can be used to find the horizontal distance travelled by the ball and the angle and magnitude of an initial velocity.

Formulae:

In kinematic equation, the distance in general notation is given by,

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

And Velocity is given by,

$v_f^2=v_0^2+2as$ (ii)

$v_f=v_0+at$ (iii)

Let’s consider an initial velocity as$v_0$and final velocity as$v_f$. So, the horizontal component of the velocity during the launch would be equal to the horizontal component of the velocity at the moment of landing.

Now, using the notation in equation (ii), the velocity in vertical direction can be written as,

$V_f^2=V_0^2+2a∆y$

Object is landed with an angle$60°$and velocity$v_f$, so,

$({\mathrm{v}}_{\mathrm{f}}\sin 60{)}^2=({\mathrm{v}}_0\mathrm{sin\theta }{)}^2-2\times 9.8\times 20$

$(v_{f}sin60{)}^2=(v_{0}sin\theta {)}^2-392$

Now, For vertical component the equation (i) can be written as,

$\Delta y=v_{0}t+\frac{1}{2}a{t}^2$

For vertical component

$20=\left({\mathrm{v}}_0\mathrm{sin\theta }\right)4-0.5\times 9.8\times 16$

$v_{0}sin\theta =24.6m/s$

Substitute this value intheabove equation,

$$\begin{gathered} {\left({\mathrm{v}}_{\mathrm{f}}\sin 60\right)}^2={\left({\mathrm{v}}_0\mathrm{sin\theta }\right)}^2-392 \\ {\left({\mathrm{v}}_{\mathrm{f}}\sin 60\right)}^2={\left(24.6\right)}^2-392 \\ {\left({\mathrm{v}}_{\mathrm{f}}\sin 60\right)}^2=-213.16 \\ {\mathrm{v}}_{\mathrm{f}}=16.8586\mathrm{m}/\mathrm{s} \\ \approx 16.9\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, using

${\mathrm{v}}_0\mathrm{cos\theta }=8.429\hspace{0.33em}\mathrm{m}/\mathrm{s}$

and

${\mathrm{v}}_0\mathrm{sin\theta }=24.6\mathrm{m}/\mathrm{s}$

Thus the horizontal distance traveled

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_0\mathrm{cos\theta }\times \mathrm{t} \\ =8.429\times 4 \\ =33.68\mathrm{m} \\ \approx 33.7\mathrm{m} \end{gathered}$$

From above two equations, taking the ratio of them,

$\tan \theta =2.9183$

Hence,

$\theta =71.1°$

Therefore,

${\mathrm{v}}_0=26.00\mathrm{m}/\mathrm{s}$

Thus, magnitude of an initial velocity is${\mathrm{v}}_0=26.00\mathrm{m}/\mathrm{s}$

From above,

$\tan \theta =2.9183$

Hence,

$\theta =71.1°$

Thus, angle relative to horizontal is $\theta =71.1°$in clockwise direction