Question

A body is thrown vertically upwards with a velocity u . calculate :
1)maximum height attained by the body ,
2)time of ascent
3)time of decent,
4)speed of yhe body on reaching the ground .​

Answer 1

Answer:

h=u*u/2

at maximum hiegh time of assent and time of descent r equal

u/g+u/g=2u/g

Explanation:

Time of Ascent:

The time taken by body thrown up  to reach maximum height h is called as time of ascent.

Let t₁ be time of ascent.

at maximim height , its velocity is zero, V=0

hence , V=u-gt

0=u-gt₁

t1=u/g ------------- (1)

Time of descent:

after reaching the maximum height, the body begins to travel downwards and reaches the ground.

The time taken to by a freely falling body to touch the ground is called time of descent.

Initial velocity for downward journey is zero  , u= 0

let t₂ be time of descent.

s=ut+1/2 gt²

h=0+1/2gt₂²

t2=√2h/g

but, h=u²/2g

so, we get

t₂=√2u²/2g

  =u/g---------------(2)

from equation 1 and 2 we can say

t₁=t₂

the time of ascent is equal to time descent in the case of bodies moving under gravity

Answer 2

OLA!!

Refer to attachment⤴⤴

tysm❤