A body of mass 0.6 kg is thrown vertically upward from the ground with a speed of 20 m/s calculate it kinetic energy just before it hit the ground
To calculate the kinetic energy just before the body hits the ground, we need to find the velocity of the body when it reaches the ground.
Using the equation of motion for vertical motion:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity = 20 m/s (given)
a = acceleration due to gravity = -9.8 m/s^2 (negative because it acts downward)
s = height = unknown
When the body hits the ground, its final displacement (s) will be equal to the height from which it was thrown up. Let's assume the initial height is zero. Therefore, s = 0.
v^2 = u^2 + 2as
v^2 = (20 m/s)^2 + 2(-9.8 m/s^2)(0)
Simplifying:
v^2 = 400 m^2/s^2 + 0
v^2 = 400 m^2/s^2
Taking the square root of both sides:
v = √(400 m^2/s^2)
v = 20 m/s
The final velocity of the body when it hits the ground is 20 m/s.
Now, we can calculate the kinetic energy just before it hits the ground using the formula:
Kinetic Energy = (1/2) * mass * velocity^2
Kinetic Energy = (1/2) * 0.6 kg * (20 m/s)^2
Kinetic Energy = (1/2) * 0.6 kg * (400 m^2/s^2)
Kinetic Energy = 0.3 kg * 400 m^2/s^2
Kinetic Energy = 120 J
Therefore, the kinetic energy just before the body hits the ground is 120 Joules.