Question

The maximum height reached by a projectile is
“h'. Its time of flight is​

Answer 1

Answer:

u^2sin^2A/2g

Explanation:

time taken in ascending the maximum height equal the time taken in descending back to the same horizontal plane

Answer 2

Explanation:

Maximum height, H=

$u {}^{2} \sin( {}^{2} ) \alpha \div 2g$

$u \sin( \alpha ) = \sqrt{2gh}$

$t = 2u \sin( \alpha ) \div g$

$2 \div g \times \sqrt{2gh}$

$= \sqrt{8h \div g}$

Hence ans is

√8h/g