A ball is thrown horizontally from the top of a building 54 m high. The ball strikes the ground at a point 35 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground

V=34.2 m/s

Explanation:

Given that

Height , h= 54 m

Horizontal distance , x = 35 m

Given that , the ball is thrown horizontally , therefore the initial vertical velocity will be zero.

In vertical direction :

We know that

Now by putting the values in the above equation we got

Assume

Thus

We also know that

In horizontal direction :

Thus the resultant velocity

V=34.2 m/s

Therefore the velocity will be 34.2 m/s.

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force