Question

question 1 (1 point)
a bullet shot straight up returns to its starting point in 10 s. what is the initial speed of the bullet?
9.8 m/s
25 m/s
49 m/s
98 m/s

Explanation:

Step1: Analyze the motion time

The total time of flight \(t = 10\ s\). The time to reach the maximum - height is half of the total time. So \(t_{up}=\frac{t}{2}=\frac{10}{2}=5\ s\).

Step2: Use the kinematic equation

At the maximum - height, the final velocity \(v = 0\). The kinematic equation \(v=v_0 - gt\) (taking upward as positive, \(g = 9.8\ m/s^{2}\)).
Substitute \(v = 0\), \(t=t_{up}=5\ s\) and \(g = 9.8\ m/s^{2}\) into \(v = v_0−gt\).
We get \(0 = v_0-9.8\times5\).

Step3: Solve for the initial velocity

Rearrange the equation \(v_0-9.8\times5 = 0\) to find \(v_0\).
\(v_0=9.8\times5=49\ m/s\).

Answer:

C. \(49\ m/s\)