Question

A ball is released from the top of the tower of height 45m. What will be the velocity while hitting the ground

Answer 1

Answer:

When the ball is at 45m height it has no kinetic energy but have some potential energy which's value is= mgh

Where m=the mass of the ball.

Now when the ball is just touch the ground it has no potential energy, all potential energy converted to kinetic energy which's value is =

$\frac{1}{2} m {v}^{2}$

Where v is the velocity at that point.

So according to the energy conservation law

Potential energy(at top) =kinetic energy (at ground)

Or, mgh=0.5(mv^2)

Or,

$v = \sqrt{2gh}$

Here h=45 m

g=9.8 m/s^2

So. V=29.7 m/s

By taking g=10m/s^2

We have

V=30m/s