A projectile is projected with an initial velocity u, at an angle \theta to the horizontal. The maximum height attained is.
A
U2 sin \(\frac{\theta}{g}\)
B
2U2 sin2 \(\frac{\theta}{g}\)
C
U sin2 \(\frac{\theta}{g}\)
D
U2 sin2\(\frac{\theta}{2g}\)
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Using V\(^2\) = (Uy)\(^2\) - 2gH
But Uy = usin\theta
At max. Height, V = 0
0 = (Uy)\(^2\) - 2gH
- (Uy)\(^2\) = - 2gH
H = \(\frac{u^2(sin)^2\theta}{2g}\)
Source: Myschool Past Questions Archive (Record #43006)