Solution: V 0 = 28 m/s, θ = 30 degrees (a) The maximum height = H max = ( V 0sinθ ) 2/(2 g) = (28 Sin 30)^2 / (2×9.8) m = 10 m (b) the time taken by the ball to return to the same level = Total time = T tot = 2(V 0sinθ )/g = (2x28xsin 30) /9.8 = 2.9 s (c) the distance from the thrower to the point where the ball returns to the same level = Horizontal range of a projectile = R = (V 0 2 sin2θ )/ g = (28x28xsin 60)/9. The range of the projectile is the total horizontal distance traveled during the flight time.

A particle is projected at a speed of u (m/s) at an angle of a to the horizontal: Range. Calculate (a) the maximum height (b) the time taken by the ball to return to the same level (c) the distance from the thrower to the point where the ball returns to the same level. The suvat equations can be adapted to solve problems involving projectiles. Projectile Motion formula | equations of projectile motion Projectile Motion Solved ExampleĪ ball is thrown at a speed of 28 m/s in a direction 30 degrees above the horizontal. In these formulas: represent the starting position. y g x 2 2 ( V 0 cos ( )) 2 + x tan ( ) The above equation is the path of projectile which is a parabola of the form y A x 2 + B x where A g 2 ( V 0 cos ( )) 2 and B tan ( ) Time of Flight of the Projectile The time of flight is the time taken for the projectile to go from point A to point C (see figure above). The initial velocity component along X-axis = V 0x = V 0 cosθ and the initial velocity component along Y-axis = V 0y = V 0sinθ. Using the displacement formula we can calculate the position ((x,y) coordinates) of a projectile at any given time.

Figure 1: Projectile motion formula or equations

Projectile formula