Question

the time of flight of projectile is 10 second and its range is 500 the maximum height reached by it will be G is equal to 10 metre per second square​

Answer 1

Answer:

h=125m

Explanation:

By equating the equations, you get ucos∝=usin∝. ∴∝=45°, i.e. angle of projection.

by equating time of flight formula, you will get usin∝=50 .....(i),

and by equating range after substituting (i) we get ucos∝=50

Hence ∝=45° and u will be calculated to 50√2 m/s.

The max height by formula comes to be= 125 m.