Question

An object is thrown from the ground at an angle of 30⁰( above horizontal) with a speed of 45.2 m/s. The total flight time of this projectile is ?
(i.e. the time it takes the object to hit the ground)

Answer 1

The initial velocity u' can be resolved in to two components

Horizontal = u' cos and vertical = u' sin, where, = 30^o

Taking motion of the object from initial point to the highest point (point of maximum height) and using v = u+at, where. u = initial velocity = u' sin, a = -g = -9.8 m/s² (-ve sign due to upward motion), v = final velocity at highest point = 0, we have, 0 = u' sin - gt => t = u' sin/g = 45.2*sin 30^o/9.8

Thus, t = 2.306 second

Now, if air resistance is neglected, then,

Time taken to go to highest point = Time taken to reach ground from highest point So, total time of flight = t+t = 2t = (2*2.306) = 4.61 second