A object 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