find the velocity with which it hits the ground when a body is projected horizontally from top of a tower 100 meter high with a velocity of 9.8 meter per second.

Vx = 9.8

Vy = √(2as) = √(2*9.8*100) = 44.27
final v = √(Vx^2 + Vy^2)

To find the velocity with which the body hits the ground, we can use the kinematic equation for vertical motion.

The equation we need to use is:

v^2 = u^2 + 2as

Where:
v = final velocity (the velocity with which the body hits the ground)
u = initial velocity in the upward direction (which is 0 m/s since the body is projected horizontally)
a = acceleration due to gravity (which is approximately 9.8 m/s^2)
s = vertical displacement (which is 100 m, the height of the tower)

Let's substitute the given values into the equation:

v^2 = 0^2 + 2 * 9.8 * 100

Simplifying:

v^2 = 0 + 1960

v^2 = 1960

Now, to find the velocity, we take the square root of both sides:

v = √1960

Calculating the square root:

v ≈ 44.27 m/s

Therefore, the velocity with which the body hits the ground is approximately 44.27 m/s.