A projectile is fired horizontally from a gun that is 45m above flat ground, emerging from the gun with speed 250m/s
How long does the projectile remain in the air? At what horizontal distance from the firing point does it strike the ground?.What is the magnitude of the vertical component of its velocity as it strikes the ground?
To solve this problem, we can make use of the equations of motion for projectile motion.
1. Time of flight:
Since the projectile is fired horizontally, the initial vertical velocity is zero. We can use the equation for vertical displacement to find the time of flight (T):
Δy = Vyi * T + (1/2) * a * T^2
0 = 0 * T + (1/2) * (-9.8) * T^2
0 = -4.9 * T^2
T^2 = 0
T = 0
Therefore, the time of flight is zero seconds.
2. Horizontal distance:
Since the projectile is fired horizontally, the initial vertical velocity is zero. The horizontal component of the velocity (Vx) remains constant. We can use the equation for horizontal displacement to find the horizontal distance (X):
Δx = Vx * T
Δx = (250 m/s) * (0 s)
Δx = 0 m
Therefore, the projectile strikes the ground exactly at the firing point, which is 0 meters horizontally from the initial position.
3. Vertical component of velocity:
Since the projectile was fired horizontally, there is no vertical component of velocity when it strikes the ground. Therefore, the magnitude of the vertical component of velocity as it strikes the ground is zero.