Home
Class 12
PHYSICS
For ground to ground projectile, time ta...

For ground to ground projectile, time take by a particle to ge point O to point C is `T_(1)` and during the same motion time taken by the particle to ge from A to B is `T_(2)` then height 'h' is :

A

`(1)/(2)gT_(1)^(2)`

B

`(1)/(4)g(T_(1)^(2)-T_(2)^(2))`

C

`(1)/(8)g(T_(1)^(2)-T_(2)^(2))`

D

`(1)/(8)g(T_(2)-T_(1))^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
For ground ……………
`T_(1) = (2V_(y1))/(g)`
`T_(2) = (2V_(y2))/(g)`
`(V_(y2))^(2) = (V_(y1))^(2) - 2gh`
Promotional Banner

Similar Questions

Explore conceptually related problems

Time taken by the particle to reach from A to B is t . Then the distance AB is equal to

Let A, B and C be points in a vertical line at height h, (4h)/(5) and (h)/(5) and from the ground. A particle released from rest from the point A reaches the ground in time T. The time taken by the particle to go from B to C is T_(BC) . Then (T)/(T_(BC)) ^(2) =

A ground to ground projectile is at point A at t = (T)/(3) , is at point B at t = (5T)/(6) and reaches the ground at t = T . The difference in heights between points A and B is

A particle is dropped from point A at a certain height from ground. It falls freely and passes through thre points B, C and D with BC=CD . The time taken by the particle to move from B to CD is 2s and from C to D is 1s . The time taken to move from A to B is s

A particle is thrown vertically upwards from the surface of the earth. Let T_(P) be the time taken by the particle to travel from a point P above the earth to its highest point and back to the point P. Similarly, let T_(Q) be the time taken by the particle to travel from another point Q above the earth to its highest point and back to the same in terms of T_(P), T_(Q) and H, is :-

A particle executes a simple harmonic motion of time period T. Find the time taken by the particle to go directly from its mean position to half the amplitude.

A particle startsits SHM at t=0 . At a particular instant on its way to extreme position , x= (A)/(2) . Find the time taken by the particle to come back to this point after passing through the extreme position. Hint : t_(1) = time taken to go from x=0 to x= (A)/(2) t_(2) = time taken to go to extreme position and come back to mean position = (T)/(2) Required time = (T)/(2) - t_(1)

In ground to ground projectile motion under gravity, which of the following doesn't affect the time of flight ?

A particle is projected vertically upwards from a point A on the ground. It takes t_(1) time to reach a point B but it still continues to move up. If it takes further t_(2) time to reach the ground from point B then height of point B from the ground is :-

Find the time taken by the particle in going from = x = (A)/2) to x=A Hint : t_(1) = Time taken to go from x=0 to x =A t_(2) = Time taken to go from x=0 to x= (A)/(2) Required time = t_(1) - t_(2)