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22 March, 12:16

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

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  1. 22 March, 13:44
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    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
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