In Fig 4-37, a ball is thrown leftward from the left edge of the roof, at height h above the ground. The ball hits the ground 1.50 slater, at distance d=25.0 mfrom the building and at angle $\mathit{\theta }\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$with the horizontal. (a) Find h. (Hint: One way is to reverse the motion, as if on video.) What are the (b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown? (d) Is the angle above or below the horizontal?

The time in which the ball reaches on the ground from the roof of the building is t = 1.5 s

The horizontal distance travelled by the ball is d = 25 m

The angle made with horizontal with which the ball hits the ground is $\theta =60°$.

This problem is based on kinematic equations that describe the motion of an object with constant acceleration. Using these equations, the height of the building, the magnitude of velocity, and the angle at which the ball is thrown can be calculated.

Formulae are as follow:

Newton’s second kinematic equation is given by,

$∆y=\left(v_{0y}\right)\left(t\right)+\frac{1}{2}a{t}^2$. (i)

The horizontal displacement of the projectile

$$\begin{gathered} x=v_{0x}\times t \\ =v_0\cos \theta \times t \end{gathered}$$ (ii)

The angle is given by

$\theta ={\tan }^{-1}\left(\frac{v_y}{V_x}\right)$ (iii)

Using equation (ii), The horizontal displacement of the ball is given by,

$25=V_0\cos \left(60\right)(1.5)$

$$\begin{gathered} \begin{array}{ll}{v}_0=\frac{25}{0.75}& \end{array} \\ =33.3m/s \end{gathered}$$

Using equation (i) the height of the building (the vertical displacement) is,

$$\begin{gathered} h=\left(33.3\times \sin \left(60\right)\right)\left(1.5\right)-\frac{1}{2}\left(9.8\right){\left(1.5\right)}^2 \\ h=32.3m \end{gathered}$$

The horizontal component of the velocity at the roof is the same as that on the ground. So,

$$\begin{gathered} v_x=33.3\times cos\left(60\right) \\ =16.7m/s \end{gathered}$$

The vertical component of the velocity at the roof is different from that on the ground.

According to Newton's first kinematic equation,

$v_y=v_0+at$

The vertical component of the velocity at the roof is,

$$\begin{gathered} v_y=33\times \sin \left(60\right)-\left(9.8\right)\left(1.5\right) \\ {v}_y=14.2m/s \end{gathered}$$

The magnitude of the velocity of the ball is,

$$\begin{gathered} v=\sqrt{v_x^2+v_y^2} \\ v=\sqrt{16.7^2+14.2^2}. \\ v=21.9m/s \end{gathered}$$

The angle relative to the horizontal with which the ball is thrown is,

$\theta =ta{n}^{-1}\left(\frac{v_y}{v_x}\right)$

$\theta =ta{n}^{-1}\left(\frac{14.2}{16.7}\right)$

$\theta =44.4^0$

It is the angle measured in the downward direction.