30.6k views
4 votes
A bullet shot straight up returns to its starting point in 10 s. what is the initial speed of the bullet, assuming negligible air resistance?

2 Answers

5 votes

Final answer:

The initial speed of the bullet shot straight up can be calculated using the equations of motion and the time it takes to return to its starting point. With a total time of 10 seconds, the time to reach the peak is 5 seconds. Considering gravitational acceleration of 9.8 m/s² and final velocity at the peak of 0 m/s, the initial speed is found to be 49 m/s.

Step-by-step explanation:

To calculate the initial speed of the bullet shot straight up, we can use the equations of motion for uniformly accelerated motion (since air resistance is negligible). The bullet experiences acceleration due to gravity, which is approximately 9.8 m/s2 downwards.

When the bullet is shot upwards, it slows down until it reaches its highest point, where its velocity temporally becomes zero. Then it starts falling back until it reaches its starting point again. The total time for this journey is given as 10 seconds. Therefore, the time to reach the highest point is half of the total time, which is 5 seconds.

The equation v = u + at describes the relationship between final velocity (v), initial velocity (u), acceleration (a), and time (t). Since the final velocity at the highest point is 0 m/s, we can set v to 0 m/s and solve for u:

0 = u + (-9.8 m/s2)(5 s)

This implies that:

u = 9.8 m/s2 × 5 s = 49 m/s.

Hence, the bullet's initial speed was 49 m/s.

User Nadia
by
8.2k points
4 votes
Time to return = flight time = 10 s


Time to reach the maximum height = flight time / 2 = 10s / 2 = 5s


Speed at the maximum height = Vf = 0


Velocity formula: Vf = Vo - g*t => Vo = Vf + gt


Vf = 0
g = 9.8 m/s^2
t = 5s


Vo = 0 + 9.8 m/s^2 * 5s = 49 m/s


Answer: 49 m/s

User Boris Burkov
by
8.7k points

No related questions found