For the given values of initial velocity of projection and angle...

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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

Civil Engineering:

4 months ago

1. If (lpha ) is measured from the HORIZONTAL: The correct answer is indeed (lpha =rac{pi }{4}+rac{eta }{2}). Reasoning: To maximize range up an incline, you must "split the difference" between the vertical ((90^{circ })) and the incline itself ((eta )). The angle halfway between (eta ) and (90^{circ }) is (rac{90+eta }{2}=45^{circ }+rac{eta }{2}).This matches Exam Veda's Option B. 2. If (lpha ) is measured from the INCLINED PLANE: The correct answer is (lpha =rac{pi }{4}-rac{eta }{2}). Reasoning: If ( heta ) is the angle from the horizontal and (lpha ) is the angle relative to the plane, then ( heta =lpha +eta ). Substituting the horizontal condition ((45^{circ }+rac{eta }{2})) into this gives:(lpha +eta =45^{circ }+rac{eta }{2}implies lpha =45^{circ }-rac{eta }{2})This matches Option A.
Amit Kumar:

7 years ago

Please provide solution.....

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