Question

A particle is projected upwards with a velocity u .....
Find max height

Answer 1

Consider the motion from A to B:

s = +H (final point lies above the initial point), initial velocity = u, final velocity v = 0.

Let the time taken to go from A to be t

1

Using v = u - gt, we get

0 = u - gt

1

⇒ t

1

=

g

u

Using s = ut - (1/2)gt

2

,

H = u(

g

u

) -

2

1

g (

g

u

)

2

=

2g

u

2

So the maximum height attained is H =

2g

u

2

Consider the return motion from B to A:

s = -H (final point lies below the initial point)

u = 0 (at point B, velocity is zero)

Let time taken to go from B to A be t

2

. We have

t

2

=

g

2H

=

g2g

2u

2

=

g

u

Here t

1

is known as the time of ascent and t

2

is known as the time of descent. We can see that

Time of ascent = Time of descent =

g

u

Total time of flight, T = t

1

+ t

2

=

g

u

Here time of flight is the time for which the particle remains in the air.

Answer 2

(u^2)/2g

Hope it helps mate