Question

6 a student throws a tennis ball straight up and catches it at the same height, on the way down. if the ball is in the air for 3.2 seconds, then what was the initial velocity of the ball? formula 1 point answer

Explanation:

Step1: Consider the motion symmetry

The time - up is equal to the time - down. The total time of flight $t = 3.2\ s$, so the time to reach the maximum height $t_{up}=\frac{t}{2}=\frac{3.2}{2}=1.6\ s$.

Step2: Use the kinematic equation

The kinematic equation for vertical motion $v = v_0−gt$. At the maximum height, $v = 0$. We want to find the initial velocity $v_0$. Rearranging the equation for $v_0$ gives $v_0=v + gt$. Substituting $v = 0$, $g = 9.8\ m/s^2$ and $t = 1.6\ s$ into the equation, we get $v_0=0+9.8\times1.6$.

Step3: Calculate the initial velocity

$v_0=9.8\times1.6 = 15.68\ m/s$.

Answer:

$15.68\ m/s$