Question

for constant acceleration conditions: (d = d_0 + v_0t+\frac{1}{2}at^{2}) (v^{2}=v_0^{2}+2a(d - d_0)) (v = v_0+at) use (g = 10.0 m/s^{2}) for the acceleration due to gravity near the surface of the earth question 3 (4 points) a basketball is tossed vertically up into the air from the edge of a roof 12.2 m above the ground. first, if the ball took 2.5 seconds from the point when released upward until it hit the ground, what was the balls initial speed? ignore air resistance and use a value of (g = 10.0 m/s^{2}) when answering this question. your answer:

Explanation:

Step1: Identify the kinematic - equation

We use the equation $d = d_0+v_0t+\frac{1}{2}at^2$. Here, $d = 0$ (ground - level), $d_0=12.2$ m (initial height), $t = 2.5$ s, and $a=-g=- 10.0$ m/s².
So, $0 = 12.2+v_0\times2.5-\frac{1}{2}\times10\times2.5^2$.

Step2: Simplify the equation

First, calculate $\frac{1}{2}\times10\times2.5^2 = 5\times6.25 = 31.25$.
The equation becomes $0 = 12.2 + 2.5v_0-31.25$.

Step3: Rearrange the equation to solve for $v_0$

Rearrange to get $2.5v_0=31.25 - 12.2$.
$2.5v_0 = 19.05$.

Step4: Solve for $v_0$

$v_0=\frac{19.05}{2.5}=7.62$ m/s.

Answer:

$7.62$ m/s