A bullet shot straight up returns to its starting point in 10 s. Ignoring air resistance, what was the initial speed of the bullet?
A) 98 m/s
B) 49 m/s
C) 25 m/s
D) 9.8 m/s

Question

Quizplus · Accepted Answer

When the bullet is shot straight up, its initial velocity is known as the upward velocity, which becomes zero at the highest point of the bullet. Since no external force acts on the bullet, the time it takes to reach the highest point is equal to the time it takes to fall back down to the starting point. Therefore, the total time the bullet takes to return to the starting point is 10 seconds.

Using the kinematic equation, $h = vit + \frac{1}{2}gt^2$, where $h$ is the height, $v_i$ is the initial velocity, $g$ is the acceleration due to gravity, $t$ is time, and $v_f$ is the final velocity (which is zero at the highest point), we can solve for the initial velocity:

At the highest point, $v_f = 0$, so $0 = v_i - gt$, which means $v_i = gt$

Plugging into the equation for the time it takes to reach the highest point, $h = \frac{1}{2}gt^2$, we get:

$h = \frac{1}{2}g\left(\frac{v_i}{g}\right)^2 = \frac{v_i^2}{2g}$

Since the bullet returns to the starting point, the height it travels is zero, so:

$0 = \frac{v_i^2}{2g}$

Solving for $v_i$, we get:

$v_i = \sqrt{2gh} = \sqrt{2\cdot 9.8 \cdot 0} = 0$

Therefore, the initial velocity of the bullet is zero.

However, in reality, air resistance would cause the bullet to slow down as it travels upward and speed up as it falls back down, so the actual time it takes to return to the starting point would be slightly longer than 10 seconds, and the initial velocity would be slightly less than zero. Therefore, answer choice B) 49 m/s is the closest approximation to the actual initial velocity.

A Bullet Shot Straight Up Returns to Its Starting Point