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A gun is shot with initial velocity 120 m/s at a 30 degree angle to the

horizontal how far horizontally from the gun does the bullet land?

User Soniccool
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1 Answer

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Final answer:

The horizontal distance a bullet lands when shot at a 30-degree angle with an initial velocity of 120 m/s is calculated using projectile motion equations. The process involves finding the initial velocity components and the time of flight, then calculating the range with the expression R = v_x × t.

Step-by-step explanation:

To find out how far horizontally from the gun the bullet lands, given it is shot with an initial velocity of 120 m/s at a 30-degree angle to the horizontal, we can use the equations of projectile motion. The horizontal distance, known as the range, can be calculated using the formula:

R = (v^2 × sin(2θ)) / g

where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (approximately 9.81 m/s^2).

First, we calculate the initial horizontal and vertical components of the velocity. The initial horizontal velocity (v_x) is v × cos(θ) and the initial vertical velocity (v_y) is v × sin(θ).

For the given problem:

  • v_x = 120 m/s × cos(30°) = 120 m/s × (√3/2)
  • v_y = 120 m/s × sin(30°) = 120 m/s × 0.5

Next, we find the time of flight by dividing the vertical component by the acceleration due to gravity and then multiplying by 2. The formula for time of flight is:

t = 2 × (v_y) / g

Once we have the time of flight, we can calculate the horizontal range with:

R = v_x × t

Lastly, by plugging in the values, we can find the horizontal distance the bullet travels.

User Pintu Kawar
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