Question

a stone projected vertically up with velocity v from the top of a tower reaches the ground with velocity 2v. The height of the tower is ​

Answer 1

Answer:

Explanation:

v^2-u^2=2as

here 2v is final velocity

and v is intial velocity

a=acceleration=g(in this case)

4v^2-v^2=20s

3v^2/20=s

Answer 2

Explanation:

According to 3rd kinematical eqn

v

${v}^{2} = {u}^{2} + 2as$

here a=-g as dispacement is in opposite direction of g and stone was initially at rest so u=0, s is displacement from top of tower

v^2=-2gs

s=v^2/2g

Now when stone reaches ground its final velocity is 2v

Again,

${v}^{2} = {u}^{2} + 2gs$

here, put v=2v u=0(at maximum height stone could reach its velocity at that point initial velocity is 0) s=S

4v^2 = 2gS

S=4v^2/2g

S=2v^2/g

Dispacement done by stone= dispacement done by stone from top of tower+height of tower

S= s + h

h= 2v^2/g - (-v^2/2g)

h= 2v^2/g + v^2/2g

h= 4v^2/2g + v^2/2g

h=5v^2/2g is height of tower