Question

An African Swallow, flying with a horizontal velocity of 80 m/s due North, drops a coconut from a height of 8 m. When does the coconut hit the ground after it is released?

Answer 1

Using 2nd kinematic equation in vertical direction:

h = V0y*t + (1/2)*ay*t^2

h = vertical height of coconut = -8 m

V0y = Initial vertical velocity = 0 m/sec, since Swallow is traveling in horizontal direction only

ay = acceleration in vertical direction = -g = -9.81 m/sec^2

So,

-8 = 0*t + (1/2)*(-9.81)*t^2

t = sqrt (2*8/9.81)

t = 1.28 sec = time taken by coconut to reach ground

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