

MCQOPTIONS
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1. |
A particle is projected with velocity u at an inclination θ with the horizontal. Then Maximum height (H) attained is |
A. | \(\frac{u^2 sin^2\theta}{g}\) |
B. | \(\frac{2u^2 sin^2\theta}{g}\) |
C. | \(\frac{u^2 sin^2\theta}{2g}\) |
D. | \(\frac{u^2 sin2\theta}{g}\) |
Answer» D. \(\frac{u^2 sin2\theta}{g}\) | |