A projectile is released with a speed u at an angle (theta) to the horizontal. With the aid of a diagram, show that the time of flight is equal to (frac{2uSintheta}{g}), where g is the acceleration of free fall.
Explanation
Vertical component velocity Vy = Usin(theta)
Time to reach maximium height = t
V = USin(theta) — gt
But V at maximum height = O
.-. O = Usin(theta) – gt
Hence, t = (frac{U sin theta}{g})
But time for a projectile to reach maximum he is equal to time to return to projection plane
Time of flight T = (frac{2Usin theta}{g})