Question

Sue and Len throw identical rocks off a tall building at the same time. The ground near the building is flat. Sue throws her rock straight downward. Len throws his rock downward and outward such that the angle between the initial velocity of the rock and the horizon is 30°. Len throws the rock with a speed twice that of Sue's rock. If air resistance is negligible, which rock hits the ground first?
1 : They hit at the same time.
2 : Sue's rock hits first.
3 : Len's rock hits first.
4 : It is impossible to know from the information given

Answer 1

The correct statement will be 1. i.e "They hit at the same time."

Reason:

Let's height of the building be $H$ ,speed of Sue's rock is $u$ ,thus speed of Len's rock will be $2u$ .

Sue's Case:

let $t_1$ be the time to reach Sue's rock,thus

$$H=ut_1+\frac{1}{2}gt_1^2 \hspace{1cm}(1)$$

Len's Case:

Since, Len's throws the rock obliquely with velocity $2u$ at a horizontal angle $30^{\circ}$ downward hence, the vertical component of velocity will be $2u\sin(30^{\circ})$ . Now, consider at time $t_2$ rock will reach to the ground, thus

$$H=2u\sin(30^{\circ})t_2+\frac{1}{2}gt_2^2\\ \implies H=2u(\frac{1}{2})t_2+\frac{1}{2}gt_2^2\\ \implies H=ut_2+\frac{1}{2}gt_2^2 \hspace{1cm}(2)$$

Observe that equation $(1) \&(2)$ both are exactly same hence their solution is also same. Thus time to reach both the rock is same moreover

$$t_1=t_2$$

Hence we are done.