Projectile Motion

If a ball is thrown at angle θ, vertically, the projectile undergoes free fall under gravity.Horizontally the projectile travels with constant speed.The resultant motion is a parabolic path.

x = ucosθt

y = usinθt - ½gt²

If a ball is fired at an angle, its initial velocity must first be resolved into horizontal and vertical components. For a launch speed u at angle θ to the horizontal:horizontal component = u cosθ    vertical component = u sinθThe two components are treated separately. Ignoring air resistance, the horizontal velocity stays constant, while the vertical motion has a constant acceleration of g downwards.The time of flight depends on the vertical height and vertical speed only. Find it using the vertical motion, then use it to find the horizontal distance (range) = u cosθ × t.For example, a ball fired at 12 m s^-1 at 30° has a vertical component of 12 sin30° = 6.0 m s^-1 and a horizontal component of 12 cos30° = 10.4 m s^-1.At the highest point the vertical component of velocity is zero, but the ball is still moving horizontally.