A projectile is projected with an initial velocity u, at an angle θ to the horizontal. The maximum height attained is.
Using V\(^2\) = (Uy)\(^2\) - 2gH
But Uy = usinθ
At max. Height, V = 0
0 = (Uy)\(^2\) - 2gH
- (Uy)\(^2\) = - 2gH
H = \(\frac{u^2(sin)^2θ}{2g}\)
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