So it takes $\frac{u}{g}$ secs to go up, how long will it take to come down?(as mentioned there is no air)

longer than

No worries! We‘ve got your back. Try BYJU‘S free classes today!

shorter than because in the previous case gravity is acting against velocity and here it is acting along the velocity

No worries! We‘ve got your back. Try BYJU‘S free classes today!

equal to

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

I don't know

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
equal to

Time of ascent is equal to time of descent. Hey, think for a minute, then proceed to the next video for the solution!

Suggest Corrections

Similar questions

So it takes $\frac{u}{g}$ secs to go up, how long will it take to come down?(as mentioned there is no air)

Q. If

u

is initial velocity, then it takes

\frac{u}{g}

secs to go up and

\frac{u}{g}

secs to come down when there is no air.

It is said that there is no shield found which can stop gravity acting. But there is a force called centripetal force which acts between planets so that they don't bump into each other. So is'nt it that centripetal force is a shield to gravity?

Q. Why does an object take same time to go up and come down? Although they travel same distance but the force of attraction should be large when it comes down, so logically it should take comparatively less time to come down than to go up (Neglect air resistance)

Join BYJU'S Learning Program

Select...

So it takes ugsecs to go up, how long will it take to come down?(as mentioned there is no air)

The correct option is C equal to Time of ascent is equal to time of descent. Hey, think for a minute, then proceed to the next video for the solution!

So it takes $\frac{u}{g}$ secs to go up, how long will it take to come down?(as mentioned there is no air)

The correct option is C
equal to

Time of ascent is equal to time of descent. Hey, think for a minute, then proceed to the next video for the solution!