Home
Class 10
PHYSICS
A stone thrown vertically upwards with...

A stone thrown vertically upwards with initial velocity u reaches a height h before coming down. Show that the time taken by it to go up is the same as the time taken to come down.

Text Solution

Verified by Experts

We have, ltBrgt `v=u+at" …(1)"`
`and s=ut+(1)/(2)at^(2)" …(2)"`
`therefore s=(v-at)t+(1)/(2)at^(2)`
`=vt-at^(2)+(1)/(2)at^(2)`
`therefore s=vt-(1)/(2)at^(2)" ...(3)"`
As the stone moves upward from `ArarrB, s=AB=h, t=t_(1),`
`a=-g" (retardation),"`
u = u and v = 0
`therefore" From Eq. (3), "h=0-(1)/(2)(-g)t_(1)^(2)`
`therefore h=(1)/(2)g t_(1)^(2)" ...(4)"`
As the stone moves downward from `B rarr A, t=t_(2), u=0, s=h and a=g`
`therefore" form Eq. (2), "h=(1)/(2)g t_(2)^(2)" ....(5)"` ltBrgt From Eqs. (4) and (5), `t_(1)^(2)=t_(2)^(2)`
`therefore t_(1)=t_(2)" "(because t_(1) and t_(2)" are positive")`
we can conclude that the time taken by the stone to go up is same as the time taken to come down .
Promotional Banner

Similar Questions

Explore conceptually related problems

A stone thrown vertically upwards with initaial velocity u reaches a height 'h' before coming down.Show that the time takes to go up is same as time taken to come down

A stone is thrown vertically upwards with initial velocity of 14 m s^-1 . The maximum height it will reach is [g= 9.8 m s^-2]

A stone is thrown vertically upwards with initial velocity of 14 m s^-1 . The maximum height it will reach is [g = 9.8 m s^-2]

A stone is thorwn vertically upwards with initial velocity of 40m//s . Taking g=10m//s^2 find the maximum height and total distance covered by stone

A stone is thrown vertically upward with a velocity of 30 m//s . How high will it rise? After how much time will it return to ground ? [Take g = 10 m//s^2]

From the top of a tower of height 'H', body is thrown vertically upwards with a speed 'u'. Time taken by the body to reach the ground is '3' times the time taken by it to reach the highest point in its path. Then, the speed u is

A ball is thrown vertically upwards with a velocity of 49 m//s . Calulate (i) The maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth.

A tennis ball is thrown up and reaches a height of 4.05 m before coming down. What was its initial velocity? How much total time will it take to come down? Assume g=10m//s^(2)

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:

A tennis ball is thrownb up and reahces a height of 4.05 m before coming down.What was its initial velocity? How much total time will it take to come down? Assumeg= 10m//s^2