Split Every Projectile into an Independent Horizontal and Vertical Motion

For launch speed u at angle θ, the time of flight is T = 2 u sin θ / g, the maximum height is H = u² sin² θ / (2g) and the range is R = u² sin 2θ / g = 2 uₓ uᵧ / g. Because R depends on sin 2θ, angles θ and 90° - θ give the same range, and the range at 15° (sin 30° = 0.5) is half the range at 45° (sin 90° = 1): a 1.5 km range at 15° becomes 3 km at 45°. For complementary angles the times of flight multiply to 2R/g, so t1 t2 is proportional to R.

A bomb dropped from a plane at 1960 m flying at 600 km/h (166.7 m/s) takes 20 s to fall and lands about 3.33 km ahead. A boy jumping horizontally off a 25 m building onto a 5 m building has a 20 m drop, so the fall takes 2 s, and crossing a 4 m gap needs a speed of at least 2 m/s.

For a projectile to return to a cart moving at constant velocity, it must keep the cart's horizontal speed (so its horizontal speed relative to the cart is zero) and have the right vertical speed. With the cart travelling 80 m at 30 m/s, the flight lasts 8/3 s, so uᵧ = gT/2 = 40/3 m/s relative to the cart. A boy who wants to catch the ball he threw at speed u and angle θ must run at the horizontal component u cos θ. At the top of the trajectory the vertical velocity is zero but the acceleration is still g downward, never zero.

When a question gives a time and a height, use y = u sinθ t - (1/2) g t²: for u = 25 m/s, t = 2 s, y = 5 m and g = 10, 5 = 50 sinθ - 20, so sinθ = 1/2 and θ = 30°. For bullets fired at the same speed in all directions, the farthest point is the maximum range u²/g, so the region covered on the ground is a circle of area π (u²/g)².