Question

derive a formula for max. height reached by a body for a vertically projected body​

Answer 1

Explanation:

If a body is thrown vertically upwards with initial velocity u to a height h then there will be retardation (a=-g). So we have equations of motion as:

v=u-gt ………(1)

h=ut - 1/2gt^2 …….(2)

v^2-u^2=-2gh…….(3)

At maximum height v=0 so

From (3) we get

-u^2=-2gh

h=u^2/2g

Answer 2

AnswEr :

Consider an object moving with an initial velocity 'u' reaches a maximum height 'h' in time 't' s.Upon reaching 'h' m ,the velocity of the object changes to 'v'

According to the Sign Convention,

Acceleration due to gravity would be '-g' m/s²

From the Third Kinematic Equation,

$\sf \: v {}^{2} - {u}^{2} = 2as \\ \\ \longrightarrow \: \sf \: - {u}^{2} = - 2gh \\ \\ \longrightarrow \: \sf \: u {}^{2} = 2gh \\ \\ \longrightarrow \: \sf \: h = \dfrac{u {}^{2} }{2g}$

Maximum height reached by the object would be given by u²/2g