MCQOPTIONS
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| 1. |
A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity) |
| A. | \[\frac{4{{v}^{2}}}{5g}\] |
| B. | \[\frac{4g}{5{{v}^{2}}}\] |
| C. | \[\frac{{{v}^{2}}}{g}\] |
| D. | \[\frac{4{{v}^{2}}}{\sqrt{5}g}\] |
| Answer» B. \[\frac{4g}{5{{v}^{2}}}\] | |