Application-3 :
A body projected vertically upwards from ground is at the same height h from the ground at two instants of time t1 and t2 (both being measured from the instant of projection) Now
a) \(
h = \frac{1}
{2}gt_1 t_2
\)
b) Velocity of projection = \(
u = \frac{1}
{2}g(t_1 + t_2 )
\)
c) \(
H_{\max } = \frac{1}
{8}g(t_1 + t_2 )^2
\)
d)A body dropped from height h takes time \(
\sqrt {t_1 t_2 }
\) to reach the ground