Projectile motion problem

Question

An object is thrown with an initial velocity v0 = 21 m/s and an angle α = 60° with the horizon. At what height will its velocity make β = 30° angle with the horizon?

Answer 1

The initial vertical velocity vector is given by V_{0 vert} = 21 m/s x sin 60 = 18.2 m/s.

At the point in its trajectory when the velocity vector makes an angle of 30° with the horizon, its vertical velocity component will be 21 m/s x sin 30 = 10.5 m/s.

To find the height at which this happens, use the equation V_f² = V₀² - 2gH, where V_f = the final vertical velocity (10.5 m/s), V₀ is the initial vertical velocity (18.2 m/s), g is 9.8 m/s², and H is the height in meters.

Solving for H, H = 10.5² m²/s² - 18.2² m²/s² / -2 x 9.8 m/s²

H = 11.3 m.

Answer 2

v_x = v₀cos60° = 10.5 m/s; v_y = v₀sin60° - 9.81t = 21sin60° - 9.81t = 18.19 - 9.81t;

tanβ = v_y/v_x; tan30° = (18.19 - 9.8t)/10.5; t = (18.19 - 10.5tan30°)/9.81 = 1.2363 s

y = v₀sin60°t - gt²/2 = 21sin60°·1.2363 - 9.81·1.2363²/2 = 14.987 m