Maximum range for a bullet
A 20 g bullet is shot with an initial kinetic energy of 1200 J (presumably from the ground). What is the maximum range of the bullet if it is fired at an angle such that the maximum range is equal to the maximum height? Should I just use 45 degrees as the angle?
Answer
The formula for range of a projectile is
R = v^2 / g * sin(2θ), where v is the initial velocity
So 45 degrees does get you the maximum horizontal range, but I don't think that's what the question is asking. It says to set the range equal to the maximum height, which means you might have to solve for the optimum angle.
Try using energy to help solve this problem. If the projectile is fired from the ground, the initial energy is just the kinetic energy: 1200 J. When the bullet reaches its maximum height, its energy consists of gravitational potential energy and kinetic energy (horizontal motion only). Try writing out and setting these equal:
1200 J = mgh + 1/2 m (v_h)^2
where h is the maximum height, and v_h is the velocity at the maximum height. We can get somewhere with this equation by substituting in h = R, and noting that v_h is just the x-component of the initial velocity because it doesn't change, v_h = v cos(θ)
You can solve for the initial velocity v using the initial kinetic energy provided. Then, you should have an equation in terms of the angle. You might have to use trig identities to solve for the angle, but once you do you can then solve for the range using the original formula.
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The formula for range of a projectile is
R = v^2 / g * sin(2θ), where v is the initial velocity
So 45 degrees does get you the maximum horizontal range, but I don't think that's what the question is asking. It says to set the range equal to the maximum height, which means you might have to solve for the optimum angle.
Try using energy to help solve this problem. If the projectile is fired from the ground, the initial energy is just the kinetic energy: 1200 J. When the bullet reaches its maximum height, its energy consists of gravitational potential energy and kinetic energy (horizontal motion only). Try writing out and setting these equal:
1200 J = mgh + 1/2 m (v_h)^2
where h is the maximum height, and v_h is the velocity at the maximum height. We can get somewhere with this equation by substituting in h = R, and noting that v_h is just the x-component of the initial velocity because it doesn't change, v_h = v cos(θ)
You can solve for the initial velocity v using the initial kinetic energy provided. Then, you should have an equation in terms of the angle. You might have to use trig identities to solve for the angle, but once you do you can then solve for the range using the original formula.