Final answer:
All four statements about the bullet's vertical motion are correct. The bullet's net displacement is zero after 10 seconds, it travels a total distance of 250 m, the rate of change of velocity (acceleration due to gravity) is constant, and the bullet is fired upwards with an initial velocity of 50 m/s.
Step-by-step explanation:
The student's question is about a bullet fired vertically upwards with a given gravity value. We need to assess the statements given in the question.
- A. The net displacement of the bullet in 10s is zero.
- B. The total distance travelled by the bullet in 10s is 250 m.
- C. The rate of change of velocity with time is constant throughout the motion of the bullet.
- D. The initial velocity of the bullet is directed vertically upwards.
For statement A, since the bullet returns back to its starting point, the net displacement is indeed zero.
This is a characteristic of freely falling bodies with the only force acting being gravitation.
For statement B, to calculate the distance travelled by the bullet, we use the projectile motion principle.
The bullet takes 5 seconds to reach the peak (half the total time of flight) and will travel the same distance on its way down. Using the equation ½gt² where g is the acceleration due to gravity and t is the time, the height reached (h) is ½ × 10 m/s² × (5 s)² = 125 m.
Since the bullet travels this height twice, once going up and once coming down, the total distance is 125m + 125m = 250m.
For statement C, the acceleration due to gravity is constant, which makes the rate of change of velocity with time also constant.
For statement D, we can calculate the initial velocity (u) using the equation u = gt, where g is the acceleration due to gravity and t is the time to reach the peak.
That is u = 10 m/s² × 5 s
= 50 m/s.
The bullet is fired with an initial velocity of 50 m/s directed upwards to reach the peak in 5 seconds.