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Question: A projectile is thrown into space so as to have maximum horizontal range R. Taking the point of proj...

A projectile is thrown into space so as to have maximum horizontal range R. Taking the point of projection as origin, the co-ordinates of the point where the speed of the particle is minimum are

A

(R, R)

B

(R,R2)\left( R,\frac{R}{2} \right)

C

(R2,R4)\left( \frac{R}{2},\frac{R}{4} \right)

D

(R,R4)\left( R,\frac{R}{4} \right)

Answer

(R2,R4)\left( \frac{R}{2},\frac{R}{4} \right)

Explanation

Solution

For maximum horizontal Range θ=45\theta = 45{^\circ}

From R=4HcotθR = 4H\cot\theta= 4H [As θ = 45o, for maximum range.]

Speed of the particle will be minimum at the highest point of parabola.

So the co-ordinate of the highest point will be (R/2, R/4)