Question

5 multiple choice 1 point a stone is thrown horizontally from the top of a 25.00 - m cliff. the stone lands at a distance of 40.00 m from the edge of the cliff. what is the initial horizontal velocity of the stone? 22.05 m/s 17.70 m/s 15.80 m/s 2.260 m/s clear my selection

Explanation:

Step1: Analyze vertical motion

The stone is in free - fall vertically. The vertical displacement $y = 25.00\ m$, the initial vertical velocity $u_y=0\ m/s$, and the acceleration due to gravity $g = 9.8\ m/s^{2}$. Using the equation $y=u_y t+\frac{1}{2}gt^{2}$, since $u_y = 0$, we have $y=\frac{1}{2}gt^{2}$.
$t=\sqrt{\frac{2y}{g}}$

Step2: Calculate time of flight

Substitute $y = 25\ m$ and $g = 9.8\ m/s^{2}$ into the time - of - flight formula.
$t=\sqrt{\frac{2\times25}{9.8}}\ s\approx\sqrt{5.102}\ s\approx2.26\ s$

Step3: Analyze horizontal motion

In the horizontal direction (no acceleration, $a_x = 0$), the horizontal displacement $x = 40.00\ m$. The horizontal velocity $v_x$ is constant, and we use the formula $x = v_x t$. Then $v_x=\frac{x}{t}$.

Step4: Calculate initial horizontal velocity

Substitute $x = 40\ m$ and $t\approx2.26\ s$ into the horizontal - velocity formula.
$v_x=\frac{40}{2.26}\ m/s\approx17.70\ m/s$

Answer:

17.70 m/s (the second option)