Question

A ball is thrown from a building $10\text{m}$ tall towards another building $75\text{m}$ away at an initial velocity of $30\text{m/s}$ at an angle of ${30}^{\circ }$ with the horizontal. The height of the other building is also $10\text{m}$. Take $g=10{\text{m/s}}^2$. Choose the correct statement from the options given below:

• A. The $Y$ component of the velocity is greater than the $X$ component of the velocity.
• B. The ball touches the building $1.728m$ below its original height.
• C. The ball falls short of the building by $1.728m$.
• D. The ball flies over the other building.

Answer 1

The correct option is D The ball flies over the other building.

Considering the reference of origin
($0,0$) at the top of $1^{st}$ building.

Initial velocity of the ball $=u=30m/s$


X component of velocity $u_x=30\cos {30}^0=30\frac{\sqrt{3}}{2}=15\sqrt{3}m/s$
Y component of velocity =
$u_y=30\sin {30}^0=30\times \frac{1}{2}=15m/s$

$a_x=0,a_y=-g$

Assuming that the ball reaches the building $75m$ away, we get:
$s=ut+\frac{1}{2}a{t}^2\Rightarrow s_x=u_{x}t\Rightarrow t=\frac{75}{15\sqrt{3}}=2.88s$

Hence, the ball takes a time of $2.88\text{s}$ to cover a horizontal distance of $75\text{m}$.

The acceleration in the Y axis, $a_y=-g$

In $t=2.88\text{s}$, the vertical distance covered by the ball is given by:

$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2\Rightarrow s_y=15\times 2.88+\frac{1}{2}\times (-10)\times 2.{88}^2\Rightarrow s_y=43.2-41.472=1.728m$

Since the displacement in the Y direction is $+1.728\text{m}$, It means that the ball flies over the building.