a man hods a rifle in a horizontal postion 4 feet above the ground and shoots the bullet with an initial velocity of 2000 fps horizontally. it may be assumed that the ground is level, what will be the horizontal distance from the man to the point at which the bullet strikes the ground?
bullet fall 4 ft , g = 32 ft/s^2
4 = 16 t^2
t = sqrt (4/16) = 2/8 = 1/4 second
how far does it go horizontal in 1/4 second?
d = 2000*1/4 = 500 ft
To find the horizontal distance that the bullet travels before hitting the ground, we need to consider the motion of the bullet in two dimensions: horizontally and vertically.
The horizontal motion of the bullet is simple since it has an initial velocity of 2000 fps in that direction and no forces act horizontally to change its horizontal velocity. Therefore, the time it takes for the bullet to hit the ground will be the same as the time it takes to fall vertically.
Now, let's calculate the time it takes for the bullet to hit the ground vertically.
We can use the equation for vertical motion: h = v₀t + (1/2)gt², where:
- h is the vertical displacement (4 feet in this case, since the bullet starts 4 feet above the ground)
- v₀ is the initial vertical velocity (zero since the bullet is only moving horizontally)
- t is the time it takes for the bullet to hit the ground
- g is the acceleration due to gravity (32.2 ft/s²)
Rearranging the equation: 4 = (1/2) * 32.2 * t²
Simplifying further: t² = 0.25 * 32.2
Calculating: t ≈ √(0.25 * 32.2) ≈ 1.21 seconds
Since the horizontal distance traveled by an object in motion is given by distance = velocity * time, the horizontal distance traveled by the bullet is:
distance = 2000 fps * 1.21 seconds ≈ 2420 feet
Therefore, the horizontal distance from the man to the point where the bullet strikes the ground is approximately 2420 feet.