Home
Class 11
PHYSICS
A ball is thrown up it reaches a maximum...

A ball is thrown up it reaches a maximum height and then comes down. If `t_(1) `and `t_(2)(t_(2)gtt_(1))` are the times that the ball takes to be a particular height, then the time taken by the ball toreach the heighest point is

A

`t_(1)+t_(2)`

B

`t_(2)-t_(1)`

C

`(t_(2)-t_(1))/2`

D

`(t_(1)+t_(2))/2`

Text Solution

Verified by Experts

The correct Answer is:
D
`v=u-"gt"_(1)` 1
`-v=u-"gt"_(2)` .2
1 and 2 adding we get
`0=4u-g(t_(1)+t_(2))`
`u=g/2(t_(1)+t_(2))`…………3
Generally `v=u+at`
At highest point v=0`" "a=-g`
From eq 3 we get
`:.0=g/2(t_(1)+t_(2))-"gt"`
`:.t=((t_(1)+t_(2)))/2`
Promotional Banner

Similar Questions

Explore conceptually related problems

A stone is thrown up with a speed of 9.8ms^-1 . Find maximum height reached.

A ball thrown vertically upward reached a maximum height of 20m How much time did the ball take to reach the height 20 m?

A ball thrown vertically upward reached a maximum height of 20 m . How mch time did the ball take to reach the height 20 m ?

A ball thrown vertically upward reached a maximum height of 20m What was the velocity of the stone at the instant of throwing up?

A ball thrown vertically upward reached a maximum height of 20 m . What was the velocity of the stone at the instant of throwing up?