First slide
Projection Under uniform Acceleration
Question

The trajectory of a projectile in a vertical plane is y = ax-bx2, where a and b are constants and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are:

Moderate
Solution

y = ax-bx2

For height or y to be maximum,

dydx = 0 or a-bx = 0

or x = a2b

(i) ymax. = a(a2b)-b(a2b)2 = a24b

(ii) (dydx)x= 0 = a = tan θ0

where θ0 = angle of projection

θ0 = tan-1(a)

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