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Examveda

For the given values of initial velocity of projection and angle of inclination of the plane, the maximum range for a projectile projected upwards will be obtained, if the angle of projection is

A. $$\alpha = \frac{\pi }{4} - \frac{\beta }{2}$$

B. $$\alpha = \frac{\pi }{2} + \frac{\beta }{2}$$

C. $$\alpha = \frac{\beta }{2} - \frac{\pi }{2}$$

D. $$\alpha = \frac{\pi }{4} - \frac{\beta }{4}$$

E. $$\alpha = \frac{\pi }{2} - \frac{\beta }{2}$$

Answer: Option B


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Comments ( 1 )

  1. Amit Kumar
    Amit Kumar :
    5 years ago

    Please provide solution.....

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The resultant of two forces P and Q acting at an angle $$\theta $$, is

A. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{P}}\sin \theta $$

B. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta $$

C. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\tan \theta $$

D. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta } $$

E. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$