select the correct answer.\na ball is released from a tower at a height of 100 meters toward the roof of…

Explanation:

Step1: Calculate vertical - displacement

The vertical displacement $y$ of the ball is $y = 100 - 25=75$ meters. The vertical - motion equation is $y = v_{0y}t+\frac{1}{2}gt^{2}$, and since the ball is released (initial vertical velocity $v_{0y} = 0$), we have $y=\frac{1}{2}gt^{2}$. $75=\frac{1}{2}\times9.8t^{2}$

Step2: Solve for time $t$

$t^{2}=\frac{2\times75}{9.8}$, so $t=\sqrt{\frac{150}{9.8}}\approx\sqrt{15.31}\approx3.91$ s.

Step3: Calculate horizontal velocity

In horizontal direction (no acceleration, $a_x = 0$), the horizontal displacement $x = 20$ meters, and the equation is $x = v_x t$. We know $x = 20$ m and $t\approx3.91$ s, so $v_x=\frac{x}{t}=\frac{20}{3.91}\approx5.11$ m/s.

Answer:

E. 5.11 meters/second