Question

question 8 (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 of the bullet

When a bullet is shot straight up and returns to the starting point, the time of flight \( T \) is related to the initial velocity \( u \) by the kinematic equation for vertical motion. The time to reach the maximum height is half of the total time of flight. So the time to reach maximum height \( t=\frac{T}{2}=\frac{10}{2} = 5\space s\). At the maximum height, the final velocity \( v = 0\space m/s\). The acceleration due to gravity \( g=- 9.8\space m/s^{2}\) (negative because it acts in the opposite direction of motion).

Step2: Use the kinematic equation \( v=u + gt\)

We know \( v = 0\), \( t = 5\space s\) and \( g=-9.8\space m/s^{2}\). Substituting these values into the equation \( 0=u-9.8\times5\).

Step3: Solve for \( u \)

\( u=9.8\times5 = 49\space m/s\)

Answer:

49 m/s (corresponding to the option "49 m/s")