Question

A body is projected with a speed
(v) at an angle with the horizontal
to have maximum range. At the
highest point the velocity is:

(a) 0
(b) v
(c) v/12
(d) v v2​

Answer 1

Answer:

Let the speed with which the body is projected be v.

The angle with which the body projected is θ.

We know that, R = u²sin2θ/g.

The horizontal range is maximum when 2θ = 90, beause sin 90 = 1.

Hence, if the horizontal range is maximum, θ = 45.

As the vertical acceleration is constant that is 0, the velocity at the highest point for the horizontal component will be v cos θ.

And the velocity of the body at the highest point for the vertical component will be 0, because the body has no velocity upwards nor downwards.

Now, at highest point the velocity is : v cos θ for horizontal component.

Hence, v cos V = v cos 45 = v * 1/√2 = v/√2.