MCQOPTIONS
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MCQOPTIONS
| 1. |
The equation of trajectory of projectile is given by\[y=\frac{x}{\sqrt{3}}-\frac{\text{g}{{\text{x}}^{2}}}{20}\], where x and y are in meter. The maximum range of the projectile is |
| A. | \[\frac{8}{3}\text{ m}\] |
| B. | \[\frac{4}{3}\text{ m}\] |
| C. | \[\frac{3}{4}\text{ m}\] |
| D. | \[\frac{3}{8}\text{ m}\] |
| Answer» C. \[\frac{3}{4}\text{ m}\] | |