Question

A projectile is thrown into space so as to have the maximum possible horizontal range equal to $400m$. Taking the point of projection as the origin, the coordinates of the point where the velocity of the projectile is minimum are:

• A. $(400,100)$
• B. $(200,100)$
• C. $(400,200)$
• D. $(200,200)$

Answer 1

The correct option is B $(200,100)$

Velocity of the projectile is minimum at the highest point since it has only horizontal component at the highest point.

range $R=\frac{v^{2}sin2\theta }{g}$

to maximize $\theta$ we have maximize $sin2\theta$ hence $\theta ={45}^o$

therefore $400=\frac{v^2}{g}$

Maximum height will occur at x-coordinate $\frac{R}{2}=200m$

Maximum height will be at $theta={45}^0$ :

$H=\frac{v^2{\sin }^2\theta }{2g}=\frac{v^2}{4g}=\frac{400}{4}=100m$