Question

Q.Derive an equation for projectile?

Answer 1

as we know in projectile motion horizontal distance can be given by
$x = t \div ucosx$
where X is replaced by theta
from the above eqn

Answer 2

Time to reach the Maximum Height by a projectile

When the projectile reaches the maximum height then the velocity component along Y axis i.e. Vy becomes 0. Say the time required to reach this maximum height is tmax. The initial velocity for the motion along Y axis (as said above) is V0sinθ

Considering vertical motion along y axis:

Vy = V0sinθ  – g tmax

=> 0 = V0sinθ  – g tmax

=> tmax= (V0sinθ )/g     ……………………………1

total time of flight for a projectile:

So to reach the maximum height by the projectile the time taken is  (V0sinθ )/g

It can be proved that the projectile takes equal time [ (V0sinθ )/g] to come back to ground from its maximum height.

Therefore the total time of flight for a projectile Tmax = 2(V0sinθ )/g …………………. 2