If a projectile is projected upward from the tower of height h, making an angle θ with horizontal, then, horizontal component of initial velocity
ux = u cosθ ...........(1)
and vertical component
uy = u sinθ ........(2)
On taking downward direction positive.
uy = -u sinθ
For time of flight: In vertical direction from A to B, from second equation of motion,
On comparing this quadratic equation with ax2 + bx + c = 0, we have:
\(a = 1,\ b = -\frac{2u\sin\theta}{g};\ c = \frac{-2h}{g}\)
\(\therefore \ = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\)
On substituting the values in above relation, we can calculate the value of t.
For range: ∵ Range = horizontal velocity x time of flight
\(\therefore \ R = u\cos\theta\times t\)
