3. A bullet is fired from at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?
For the horizontal:
x = vo,x * t
where
x = horizontal distance
vo,x = initial horizontal velocity
t = time
x = 1100 m/s * 0.32 s
x = 352 m
For the vertical:
h = vo,y * t - (1/2)gt^2
where
h = height
vo,y = initial vertical velocity (which is zero)
g = acceleration due to gravity = 9.8 m/s^2
h = 0 - (1/2)(-9.8)(0.32^2)
h = 0.502 m
hope this helps~ `u`
To determine how far the bullet traveled horizontally and vertically, we need to break down the velocity into its horizontal and vertical components.
Given:
- Time of flight (t) = 0.32 seconds
- Initial velocity (v) = 1100 m/s
Step 1: Calculate the horizontal distance traveled (x):
Since the horizontal velocity remains constant throughout the projectile's flight, we can use the formula:
x = v * t
Substituting the values:
x = 1100 m/s * 0.32 s
x = 352 meters
So, the bullet traveled 352 meters horizontally.
Step 2: Calculate the vertical distance traveled (y):
We can use the formula for motion in the vertical direction:
y = (v * t) - (0.5 * g * t^2)
Where:
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the values:
y = (1100 m/s * 0.32 s) - (0.5 * 9.8 m/s^2 * (0.32 s)^2)
y = 176 meters - 0.49984 meters
y = 175.50016 meters
So, the bullet traveled 175.5 meters vertically.
Therefore, the bullet traveled 352 meters horizontally and 175.5 meters vertically in the given time.