a body is thrown vertically upward from the earth's surface and it took 8 seconds to return to its original position . find out the initial velocity with which it was thrown up

The initial velocity with which the body was thrown up is 39.2 m/s

Explanation:

The given parameters for the body are;

The time it takes the body to return back to its initial position = 8 seconds

To answer the question, we make use of the kinematic equation of motion, v = u - g·t

Where

v = The final velocity of the body = 0 m/s at the maximum height

u = The initial velocity

g = The acceleration due to gravity = 9.8 m/s²

t = The time in which the body spends in the air

Therefore, at maximum height, we have;

v = 0 = u - g·t

u = g·t

t = u/g

From h = 1/2gt², which gives t = √(2·h/g), the time the body takes to maximum height = The time the body takes to return to its original position from maximum height.

Therefore, the total time in which the body is in the air = 2 × t = 2× u/g

∴

The total time in which the body is in the air = The time it takes the body to return back to its initial position after being thrown = 2 × t =  8 seconds

∴ 2 × t = 8 s = 2 × u/g

8 s = 2 × u/g

u = (8 s × g)/2

∴ u = (8 s × 9.8 m/s²)/2 = 39.2 m/s

The initial velocity with which the body was thrown up = u = 39.2 m/s.

no. dot product   =0 when vecs at right angles

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