Equations of Motion

A footballer chips the ball over the goalkeeper; a stone is thrown horizontally off a bridge into the River Niger; a mango drops from a tree. Each is a body moving under gravity alone, and each can be solved with the same equations of motion you already know, provided you treat horizontal and vertical motion separately. In this lesson you will apply v = u + at, s = ut + ½at² and v² = u² + 2as to free fall and vertical throws, then to projectiles launched horizontally and at an angle. These are some of the most demanding mechanics questions in UTME.

For vertical motion, choose upwards as positive, so a = −g = −10 m/s². A ball thrown straight up at u slows down by 10 m/s every second, stops momentarily at the top (v = 0, but a is still −10 m/s²) and then falls back. Time to the top = u ÷ g; maximum height = u² ÷ 2g; it returns to the thrower's hand with the same speed after a total time 2u ÷ g. A body projected horizontally from a height h starts with zero vertical velocity. Its time of flight depends only on the height: h = ½gt². During that time it travels horizontally at its launch speed, so the horizontal distance is x = ut. A bullet fired horizontally and a bullet dropped from the same height hit the level ground at the same time. A projectile launched at speed u and angle θ above the horizontal has components uₓ = u cos θ (constant) and uᵧ = u sin θ (decreasing at g). At the highest point the vertical velocity is zero, but the body is still moving horizontally at u cos θ. Putting the vertical motion into the equations gives: time to the top u sin θ ÷ g, time of flight T = 2u sin θ ÷ g, maximum height H = u² sin² θ ÷ 2g and range R = uₓ × T = u² sin 2θ ÷ g. Because sin 2θ is largest (equal to 1) when 2θ = 90°, the maximum range on level ground comes from θ = 45°, giving R(max) = u² ÷ g. Complementary angles such as 30° and 60° give the same range, though the 60° shot flies higher and stays up longer.