Question

please derive out ...
sub-:physics
class11

T=2u sin theta/g
WHERE T =time period
u =initial velocity
g=gravity.....

Answer 1

LHS = RHS

GIVEN EQUATION IS CORRECT.

Answer 2

♥️Hope u like my process❤️
==========================
Formula to be used
=-=-=-=-=-=-=-=-=-=-=-=
$= > \bf \: v = u - gt \\ \\ where \: \: v = final \: \: velocity \: \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: u = initial \: \: velocity \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: g = gravity \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: t = time \: \: taken$

Here, it is - gt because the object is going up.
___________________________

For a projectile motion,
=-=-=-=-=-=-=-=-=-=-=-=-=-=
For vertical component,

=>Initial velocity (u) = $\bf u \sin(\theta)$

At maximum height,

=> Time taken is t½

=> final velocity (v) =0

So,
$= > \bf \: v = u - gt \\ \\or. \: \: 0 = u \sin( \theta ) - g t_{ \frac{1}{2} } \\ \\ or. \: \: \bf \: t_{ \frac{1}{2} } = \frac{u \sin( \theta ) }{g}$

___________________________
So,

For the total time taken in projectile (t)

$= 2 \times t_{ \frac{1}{2} } \\ \\ = 2 \times \frac{u \sin( \theta ) }{g} \\ \\ \\ \\ so.. \\ \\ = > \bf total \: \: time \: \: taken(t) = \frac{2u \sin( \theta ) }{g} \\ \\\: \: \: \: \: \: \: \: \: \: \: \: ...(proved)...$
______________________________
Hope this is ur required answer ❤️♥️

Proud to help you ❤️♥️