Aobject is projected vertically upwards from a height of 45m with a velocity 30ms^-1. find the time taken to reach the ground

The total time of the projectile, is. t = 12.43 s

Explanation:

Given data,

The height from which the object is projected, h = 45 m

The initial velocity of the projectile, v = 30 m/s

The time of flight of the projectile to attain maximum height,

                                 t₁ = u/g

                                   = 30 / 9.8

                                  = 3.06 s

The maximum height of the projectile,

                                  hₓ = u²/2g

                                       = 30²/ (2 x 9.8)

                                       = 45.92 m

To find the time taken by the projectile to reach the ground from hₓ, using the II equation of motion.

                             s = ut₂ + ½ gt₂²

                             t₂ = √(2s/g)        (∵ u = 0 at hₓ)

                                = √(2 x 45.92 / 9.8)

                               = 9.37 s

Therefore the total time of the projectile,

                                    t = t₁ + t₂

                                     t = 3.06 s + 9.37 s

                                       = 12.43 s

Hence, the total time of the projectile, is. t = 12.43 s

c. magnitude and direction