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. \(\frac{u^2 sin2\theta}{g}\) | |