Question

the time of ascent when measured from the point of projection of a body projected upwards , the

Answer 1

If body is projected vertically upwards:

Let initial velocity be u , final Velocity be zero, and acceleration be g.

$\therefore \: v = u + at$

$= > 0 = u + ( - g)t$

$= > u = gt$

$\boxed{= > t = \dfrac{u}{g} }$

If Body is projected at angle $\theta$:

Total time to complete the Projectile Trajectory be t:

$\therefore \: y = u_{y}t - \dfrac{1}{2} g {t}^{2}$

$= > \: y = u \sin( \theta) t - \dfrac{1}{2} g {t}^{2}$

$= > \: 0 = u \sin( \theta) t - \dfrac{1}{2} g {t}^{2}$

$= > \: u \sin( \theta) t = \dfrac{1}{2} g {t}^{2}$

$= > t = \dfrac{2u \sin( \theta) }{g}$

So , time for ascent will be half:

$\boxed{= > t2 = \dfrac{t}{2} = \dfrac{u \sin( \theta) }{g} }$