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) ≈ 66.4 m
approximately equals 66.4 m
B) −9.8 m/s2
negative 9.8 m per s squared
C) ≈ 0.55 s
approximately equals 0.55 s
D) There is not enough information to find the range of the bullet.
There is not enough information to find the range of the bullet.
Well, this bullet must think it's competing in the high jump with that height of 1.5 m! But don't worry, I've got the answer for you, my friend. The range of a bullet fired horizontally is determined by the time it spends in the air. Now, we know that the bullet is fired horizontally, so there will be no vertical acceleration (that's gravity, in case you were wondering!). Therefore, option B can be eliminated, sorry gravity!
To find the time it takes for the bullet to reach the ground, we need to consider its vertical motion. The equation we can use is:
Δy = v₀y * t + (1/2) * a * t²
Since the bullet is fired horizontally, its initial vertical velocity (v₀y) is 0. And because it's not being influenced by gravity horizontally, the vertical acceleration (a) is also 0. That simplifies the equation to:
Δy = 0 * t + (1/2) * 0 * t²
Which just further simplifies to:
Δy = 0
Well, well, well! It looks like the bullet thinks it can fly! So, the bullet will never hit the ground vertically, meaning it will continue its noble journey forever. There is not enough information to find the range of the bullet, so our answer is option D. Keep aiming for the stars, bullet!
To find the range of a bullet fired horizontally, we need to calculate the time it takes for the bullet to hit the ground.
Since the bullet is fired horizontally, the initial vertical velocity is 0 m/s.
Using the equation of motion, h = ut + (1/2)gt^2, where h is the vertical displacement, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time of flight.
Here, h = -1.5 m (taking downward direction as negative), u = 0 m/s, and g = -9.8 m/s^2.
-1.5 = 0*t + (1/2)(-9.8)*t^2
-1.5 = -4.9t^2
Simplifying the equation:
t^2 = 1.5/-4.9
t^2 = -0.306
This equation has no real solutions since finding the square root of a negative number is not possible.
Since we cannot find a real solution for time, there is not enough information to find the range of the bullet. So the answer is:
D) There is not enough information to find the range of the bullet.
To find the range of a bullet fired horizontally, we can use the equation:
Range = (initial velocity) x (time of flight),
Given that the bullet is fired horizontally, it means that the vertical component of its initial velocity is 0. Therefore, the only relevant velocity component is the horizontal component.
Because the bullet is fired horizontally and the vertical height is 1.5 m, we can find the time of flight using the equation:
time of flight = (2 x height) / (acceleration due to gravity).
In this case, the acceleration due to gravity is approximately 9.8 m/s^2.
Let's calculate the time of flight first:
time of flight = (2 x 1.5 m) / (9.8 m/s^2)
≈ 0.306122 s
≈ 0.31 s (rounded to two decimal places).
Now, using the initial velocity (horizontal component) of 120 m/s and the time of flight of 0.31 s, we can find the range:
Range = (initial velocity) x (time of flight)
= 120 m/s x 0.31 s
≈ 37.2 m
≈ 37 m (rounded to the nearest whole number).
Therefore, the range of the bullet fired horizontally at a height of 1.5 m with an initial velocity of 120 m/s is approximately 37 meters.
The correct answer is not listed among the options provided.