## Question

It is possible to project a particle with a given velocity in two possible ways so as to make it pass through a point P at a distant *r* from the point of projection. The product of the times taken to reach this point in the two possible ways is then proportional to

### Solution

*r*

The range of a projectile is the two possible angles of projection are θ and . The time of flight corresponding to these two angles are

Thus *t*_{1}*t*_{2} ∝ *r*.

#### SIMILAR QUESTIONS

At what angle (θ) with the horizontal should a body be projected so that its horizontal range equals the maximum height it attains?

A body is projected horizontally from a point above the ground. The motion of the body is described by the equations

*x* = 2*t*

and *y* = 5 *t*^{2}

Where *x* and *y* are the horizontal and vertical displacements (in m) respectively at time *t*. The trajectory of the body is

A body is projected at time *t* = 0 from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacement *x* and *y* (in metres) respectively vary with time *t*

(in seconds) as

What is the magnitude and direction of the velocity with which the body is projected?