A bullet is fired horizontally from a height of 45 m and hits the ground 2000 m away. With what velocity does the bullet leave the fun?
To solve this problem, we can use the principles of projectile motion. When a bullet is fired horizontally, its initial vertical velocity component is zero, and its horizontal velocity component remains constant throughout its motion.
Given:
Initial vertical velocity component (Vy) = 0 m/s
Initial vertical position (y) = 45 m
Horizontal displacement (x) = 2000 m
We can use the following equation to find the time of flight (t):
y = Vy * t + (1/2) * g * t^2
Since Vy = 0, the equation simplifies to:
y = (1/2) * g * t^2
Rearranging the equation, we get:
t = sqrt((2 * y) / g)
where g is the acceleration due to gravity and its approximate value is 9.8 m/s^2.
t = sqrt((2 * 45) / 9.8)
t = sqrt(90 / 9.8)
t = sqrt(9.1836)
t ≈ 3.03 seconds
Now that we have the time of flight, we can find the horizontal velocity (Vx) using the equation:
x = Vx * t
Rearranging the equation, we get:
Vx = x / t
Vx = 2000 m / 3.03 s
Vx ≈ 660.07 m/s
Therefore, the bullet leaves the gun with an approximate horizontal velocity of 660.07 m/s.
To calculate the velocity at which the bullet leaves the gun, we can use the principles of projectile motion.
First, we need to determine the time it takes for the bullet to hit the ground. We can use the formula:
y = ut + (1/2)gt^2
Where:
y = initial height = 45 m
u = initial vertical velocity (which is 0 since the bullet is fired horizontally)
g = acceleration due to gravity = 9.8 m/s^2
t = time taken to hit the ground
By substituting the values into the formula, we get:
45 = 0t + (1/2)(9.8)t^2
45 = 4.9t^2
t^2 = 45/4.9
t ≈ 3.43 s (rounded to two decimal places)
Now that we know the time it takes for the bullet to hit the ground, we can calculate the horizontal velocity (Vx). The horizontal distance traveled (x) is given as 2000 m.
x = Vx * t
Rearranging the formula to solve for Vx:
Vx = x / t
Vx = 2000 / 3.43
Vx ≈ 581.97 m/s (rounded to two decimal places)
Therefore, the velocity at which the bullet leaves the gun is approximately 581.97 m/s.
how long does it take to fall 45m? 4.9t^2 = 45
Use that value of t:
speed = distance/time = 2000/t m/s
The horizontal speed does not change as the bullet falls.