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
I think it’s not enough info (D) because I don’t know how to solve this
Can someone pls help me. I’m probably wrong
C. 66.4 m
Because the horizontal acceleration is 0, use x = v(Δt) to find horizontal displacement (the range)
v = 120 m/s
Use t = (sqrt)Δy/(0.5[-9.81]) to find Δt
Δy = -1.5m (because it is *falling* from 1.5 meters, thus it is negative)
t = (sqrt)-1.5/(0.5[-9.81])
t = (sqrt)-1.5/-4.905
t = (sqrt)0.3058103
t = 0.553
Now substitute
x = v(Δt)
x = 120(0.553)
x = 66.36
66.36 is about 66.4
c:
To find the range of a bullet fired horizontally, you can use the equation for horizontal motion:
range = initial velocity * time
Given that the initial velocity is 120 m/s, we need to determine the time it takes for the bullet to hit the ground at a height of 1.5 m. To do that, we can use the equation for vertical motion:
final position = initial position + initial velocity * time + (1/2) * acceleration * time^2
Since the bullet is fired horizontally, the initial vertical velocity is 0 m/s. The initial position is 1.5 m and the final position is 0 m (at ground level). The gravitational acceleration is -9.8 m/s^2.
0 = 1.5 + 0 * time + (1/2) * (-9.8) * time^2
Simplifying the equation:
0 = 1.5 - 4.9 * time^2
Rearranging the equation:
4.9 * time^2 = 1.5
Dividing both sides by 4.9:
time^2 = 1.5 / 4.9
Taking the square root of both sides:
time = sqrt(1.5 / 4.9) ≈ 0.17 seconds
Now that we have the time, we can find the range:
range = initial velocity * time
range = 120 * 0.17
range ≈ 20.4 meters
So, the correct answer is C) about 20.4 m, not D) not enough info.
To solve this problem, you can use the equations of motion for horizontal projectile motion. Assuming there is no air resistance, the range of a bullet fired horizontally can be calculated using the formula:
Range = Initial Velocity × Time of Flight
To find the time of flight, we can use the equation:
Time of Flight = 2 × (Vertical Displacement) / Gravity
Since the bullet is fired horizontally, the vertical displacement is the initial height of 1.5 m, and the acceleration due to gravity is approximately 9.8 m/s^2.
Substituting these values into the equations, we can calculate the range.
Range = 120 m/s × [2 × (1.5 m) / 9.8 m/s^2]
= 120 m/s × 0.3061224 s
≈ 36.734688 m
Therefore, the correct answer is C) about 66.4 m, based on the range calculated.