Questions
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
detailed solution
Correct option is C
y=ax-bx2; for height or y to be maximum: dydx=0 or a-2bx=0 or x=a2b ymax=aa2 b-ba2 b2=a24 b dydxx-0=a=tanθ0, where θ0= angle of projection θ0=tan-1(a)Similar Questions
Assertion : When the velocity of projection of a body is made n times, its time of flight becomes n times.
Reason : Range of projectile does not depend on the initial velocity of a body.
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