What is the range of a bullet fired horizontally at a height of 1.5 m with an initial velocity of 120 m/s
A) -9.8 m/s^2
B) about 0.55 s
C) about 66.4 m
D) not enough info
Well, well, well, we have a bullet on the loose! Let's see how far this feisty fella can go. Assuming no air resistance and neglecting any deviations due to the Earth's rotation, the horizontal range of a bullet fired at an initial velocity of 120 m/s can be calculated using some good ol' physics.
The range (R) of a projectile fired horizontally can be given by the equation:
R = (Vx * t)
Where Vx is the horizontal component of the initial velocity and t is the time of flight.
Now, since the bullet is fired horizontally, its initial vertical velocity is 0 m/s. This means that the bullet won't experience any acceleration in the vertical direction due to gravity. So, the time of flight (t) can be found by dividing the initial vertical displacement (1.5 m) by the acceleration due to gravity (-9.8 m/s^2, thanks Mr. Newton).
Using the equation:
H = (V0y * t) + (0.5 * g * t^2)
Where H is the initial vertical displacement (1.5 m), V0y is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight.
Rearranging the equation to solve for t:
t = sqrt((2 * H) / g)
Plugging in the values:
t = sqrt((2 * 1.5 m) / (-9.8 m/s^2))
After crunching the numbers, we get t ≈ 0.55 s.
Now that we have the time of flight, we can calculate the range using the first equation mentioned:
R = (Vx * t)
Plugging in the values:
R = (120 m/s * 0.55 s)
And after more number crunching, we get R ≈ 66.4 m.
So, the correct answer is C) about 66.4 m. That bullet sure knows how to travel, huh?
To find the range of a bullet fired horizontally, we need to find the time it takes for the bullet to hit the ground and then multiply that time by the horizontal velocity.
Given:
Initial velocity (u) = 120 m/s
Height (h) = 1.5 m
Acceleration due to gravity (g) = 9.8 m/s^2
Step 1: Calculate the time it takes for the bullet to hit the ground.
Since the bullet is fired horizontally, the only force acting on it vertically is gravity. Therefore, we can use the equation:
h = (1/2) * g * t^2
Re-arranging the equation to solve for time:
t = √(2h / g)
Plugging in the values:
t = √(2 * 1.5 / 9.8)
t ≈ √(0.30612245)
t ≈ 0.553 s (approximately)
Step 2: Calculate the range.
The range of a projectile fired horizontally is given by the equation:
Range = u * t
Plugging in the values:
Range = 120 * 0.553
Range ≈ 66.4 m (approximately)
Therefore, the answer is C) about 66.4 m.
To calculate the range of a projectile fired horizontally, you need to consider the horizontal component of its initial velocity and the time it takes for the projectile to fall back to the ground.
In this case, the bullet is fired horizontally, so the initial vertical velocity is zero. However, the horizontal velocity remains constant due to the absence of any horizontal forces acting on the bullet.
To find the time it takes for the bullet to fall back to the ground, you can use the formula for the time of flight of a projectile:
time = sqrt((2 * height) / g)
where height is the initial vertical height of the projectile and g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Substituting the values given in the question, we have:
time = sqrt((2 * 1.5) / 9.8) = sqrt(3 / 9.8) ≈ 0.55 s
Since the horizontal velocity is constant and there is no vertical acceleration, the range (horizontal distance traveled) of the bullet is simply given by multiplying the horizontal velocity by the time of flight:
range = horizontal velocity * time = 120 m/s * 0.55 s ≈ 66.4 m
Therefore, the correct answer is C) about 66.4 m, as given.