Maximum height of a fired gun

Question

An 8 g bullet leaves a gun at 700 m s^-1.

a) What is the maximum height that this bullet could reach (ignoring air resistance)

b) If the gun is aimed at an angle of 30° above the horizontal what height will the bullet reach (again ignoring air resistance)

Answer 1

Use the kinematic equation v_f² = v_i² + 2aΔy where v_f = 0 (highest point the bullet reaches would be when it instantaneously stops to begin its fall back to earth), v_i = 700m/s and a = the acceleration of gravity or -9.81 m/s²

So, solving for Δy we get Δy = (v_f² - v_i²)/(2g) = (0 - 700²)/[2*(-9.81)] = -490000/(-19.62) = 24,974 m but to 1 sig fig it would be 20,000 m

If the gun is aimed at 30° then we need a little trig to determine what part of the 700 m/s is in the y direction.

cos(30°) = adjacent/hypotenuse = v_ix/v_i = v_ix/700 so v_ix = 700cos(30°) = 606.2 m/s

sin(30°) = opposite/hypotenuse = v_iy/v_i = v_iy/700 so v_iy = 700cos(30°) = 350 m/s

So, since in the y direction, the velocity is only 350 m/s, we use that in the kinematic equation. Δy = (v_f² - v_i²)/(2g) = (0 - 350²)/[2*(-9.81)] = -122500/(-19.62) = 6244 m but to 1 sig fig it would be 6,000 m