Question

A stone is released from the top of a tower of height 50m.Calculate the velocity of the stone just before touching the ground.

*This is a question which came for class IX periodic test*

Answer 1

here is ur answer :

1. given : mass of object = m  ,     height = 50 m ,  g (gravity) = 9.8 m/s

2.to find :  velocity of stone ?

3. solution :  we know that velocity = kinetic energy  =  1÷2 m v×v

therefore for finding the velocity of the stone , use the formula for K.E.

  but this example is the example of free fall. and in the free fall the kinetic energy is always zero .  that is why the velocity of the stone is zero.

but we can find the potential energy of the stone by using the formula

mass  ×  gravity  × height .  

hope this will help you . plz mark me as brainliest

Answer 2

_/\_Hello mate__here is your answer--

_____________________________

u = 0 m/s

v = ?

s = Height of the stone = 50 m

g = 10 ms−2

According to the equation of motion under gravity

v^2 − u^2 = 2gs

⇒ v^2 − 0^2 = 2 × 10 × 50

⇒ v^2 = 2 × 500 = 1000

⇒ v = 10√10 ms−1

Hence, the velocity of the stone just before touching the ground is 10√10 ms^−1.

I hope, this will help you.☺

Thank you______❤

__________________________❤