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Monday, August 11, 2008

Q) The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to flow in the direction of motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same condition of projection , find the horizontal range of the projectile.

A)
The time taken by it to fall down to the ground will still remain the same, just that it will cover more distance due to additional horizontal acceleration:
Along horizontal axis, initial velocity = u.cos@
acceleration along same direction = (+)g/2
Time taken by projectile (to reach ground) = 2usin@/g
s = ut + 1/2.at^2
s = ucos@.2usin@/g + 1/2*g/2*4u^2sin^2@ / g^2
s = u^2.sin2@/g + u^2.sin^2@/g
s = (u^2 / g) [sin2@ + sin^2 @]


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