A particle is thrown vertically upwards at a particular speed from the top of a tower. It reaches the ground in $t_1$ seconds. When thrown vertically downwards with the same speed, it reaches the ground in $t_2$ seconds. If it is allowed to fall freely, then the time it takes to reach the ground is given by?

How is it $\sqrt{t_1 t_2}$?

Question

A particle is thrown vertically upwards at a particular speed from the top of a tower. It reaches the ground in $t_1$ seconds. When thrown vertically downwards with the same speed, it reaches the ground in $t_2$ seconds. If it is allowed to fall freely, then the time it takes to reach the ground is given by?

How is it $\sqrt{t_1 t_2}$?

ganeshie8 · Answer

interesting, lets try by setting up equations for all 3 cases

parthkohli · Answer

I was given a hint that $t_1 = t_2 + u/g$. I understood that but nothing more.

parthkohli · Answer

I'm on a tablet so I better doodle it.

ganeshie8 · Answer

should be easy to figure out, let me grab a pen/paper

parthkohli · Answer

Hey, I'd start working at it tomorrow... I mean today. If you could post the solution here, I would be really appreciate it. Or a hint would work as well. Thanks!

anonymous · Answer

attachment

parthkohli · Answer

Oh yes! I wrote u/g instead of 2u/g. Thanks!