Question

ahmed simultaneously throws two stones from the top edge of a building with a speed 20.5 m/s. they throw one straight down and the other straight up. the first one hits the street in a time $t_1$. how much later is it before the second stone hits? $delta t =$ number (rtol=0.05, atol=1e - 08) s

Explanation:

Step1: Analyze motion of upward - thrown stone

The stone thrown upward will go up, reach its maximum height, and then fall back down to the starting - point (the top of the building) with the same speed it was thrown upward, but in the downward direction. The time it takes for the stone thrown upward to return to the starting - point is given by the kinematic equation $v = v_0+at$. When the stone reaches its maximum height, its final velocity $v = 0$. The acceleration due to gravity $a=-g=- 9.8\ m/s^{2}$ (taking downwards as positive) and the initial velocity $v_0$ is the upward - throwing velocity. Using $v = v_0+at$, for the upward motion to the maximum height, $0 = v_0 - gt_{up}$, so $t_{up}=\frac{v_0}{g}$. The total time for the stone thrown upward to return to the starting - point is $t = 2t_{up}=\frac{2v_0}{g}$.

Step2: Calculate the time difference

The time difference $\Delta t$ between the stone thrown downward and the stone thrown upward hitting the ground is equal to the time it takes for the stone thrown upward to go up and come back to the starting - point. Given $v_0 = 20.5\ m/s$ and $g = 9.8\ m/s^{2}$, we substitute into the formula $\Delta t=\frac{2v_0}{g}$.
$\Delta t=\frac{2\times20.5}{9.8}\ s$
$\Delta t=\frac{41}{9.8}\ s\approx4.18\ s$

Answer:

$4.18\ s$