Answer:
The initial velocity with which the body was thrown up is 39.2 m/s
Step-by-step 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.