Question

A boy standing at the top of a tower 180 m height drops a stone assuming g = 10 m/s2 the velocity with which it hits the ground is

Answer 1

Given that, a boy standing at the top of a tower of height 180 m drop a stone; acceleration due to gravity is 10 m/s².

We have to find that with what velocity the stone hit the ground.

From above data we have; height i.e. h of the tower is 180 m, acceleration due to gravity i.e. g is 10 m/s² and initial velocity i.e. u of the stone is 0 m/s.

First equation of motion:

v = u + at

Second equation of motion:

s = ut + 1/2 at²

Third equation of motion:

v² - u² = 2as

We have values of v, u and s. So,

Using the Third Equation Of Motion,

v² - u² = 2as

Substitute the values,

→ v² - (0)² = 2(10)(180)

→ v² = 3600

→ v = √3600

→ v = 60

Therefore, the stone hit the ground with a velocity of 60 m/s.

Answer 2

Given ,

Height , s = 180 m

Acceleration , a = Gravity = 10 m/s² [ Here ]

Initial velocity , u = 0 m/s

Final velocity , v = ? m/s

Apply 3 rd equation of motion.

$\bold{\bf{\red{v^2-u^2=2as}}}\\\\\implies \bold{v^2-0^2=2*10*180}\\\\\implies \bold{v^2=3600}\\\\\implies \sf{\bf{\blue{v=60\ m/s}}}$

So stone hits the ground with the velocity of 60 m/s.