Question

A ball is thrown vertically upwards it was observed at height h twice with a time interval t the initial velocity is

Answer 1

it's too long but.... hope you'll understand that ... :)

Answer 2

Answer:

Explanation:t1 is the time the particle takes to reach h.

so,

             h = ut1 - ½ gt12           \Rightarrow         u = (h + ½ gt12) / t1      (equation 1)

t2 is the time taken by the particle to reach the same point again.

so,

            h = ut2 - ½ gt22

  substituting 1 in the above equation:

           h = (h + ½ gt12) t2/t1 – ½ gt22

  solving

           h = ½ gt1t2    

or

Let v be initial velocity of vertical projection and t be the time taken by the body to reach a height h from ground.

Here, u=u,a=−g,s=h,t=t

Using, s=ut+21​at2, we have

h=ut+21​(−g)t2 or gt2−2ut+2h=0

∴t=2g2u±4u2−4g×2h

​​=gu±u2−2gh

​​

It means t has two values, i.e.,

t1​=gu+u2−2gh

​​

t2​=gu−u2−2gh

​​

t1​+t2​=g2u​ or u=2g(t1​+t2​)​.

or

a body is accelerating and decceleration this mean they indirectly given us time of flight

let time os ascent is t1

time of descent is t2

t1+t2=2u/g

u=g(t1+t2)/2