An 8g bullet leaves a gun at 700ms-1.
(a) what is the maximum height that this bullet could reach (ignoring air resistance)
To find the maximum height that the bullet could reach, we can use the concept of projectile motion.
We know that the bullet leaves the gun with an initial velocity of 700 m/s. Since we are ignoring air resistance, the only force acting on the bullet is the force due to gravity, which causes it to move in a parabolic trajectory.
The key equation we can use is the vertical motion equation for projectile motion:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity (0 m/s at the maximum height, since the bullet momentarily stops before falling back down)
- vi is the initial velocity (700 m/s)
- a is the acceleration due to gravity (-9.8 m/s^2, since gravity pulls objects downward)
- d is the displacement (the maximum height we want to find)
Using this equation, we can solve for d:
0^2 = (700 m/s)^2 + 2(-9.8 m/s^2)d
Rearrange the equation:
0 = 490,000 m^2/s^2 - 19.6 m/s^2 d
Solving for d:
19.6 m/s^2 d = 490,000 m^2/s^2
d = 490,000 m^2/s^2 / 19.6 m/s^2
d ≈ 25,000 m
Therefore, the maximum height that the bullet could reach, ignoring air resistance, is approximately 25,000 meters.