#### Sample Problem: Projectile

A particle is projected at an angle *q* to a horizontal plane as shown in Fig. 1. If *u* is the initial velocity of projection and we assume there is no air resistance, find expressions for:

#### Solution

Fig. 1 illustrates the physical model of the projectile’s motion and provides the basis for constructing equations whose solution makes it possible to answer the individual parts (a) ?(f):

The components of velocity are shown in Fig. 1 where the *x*-axis represents the horizontal plane and the *y*-axis, the vertical plane. We can use the format of equations derived in Section 1: “Velocity and Acceleration”:

Where:

*v* = velocity at any time *t*

*u* = Initial velocity

*a* = Acceleration = –*g*

*s = *distance travelled in time *t*

Consider the following equations of motion:

(i)* v* = *u* + *at* (ii)* v*^{2} = *u*^{2} + 2*as* (iii) *s* = *ut* +*at*^{2}

(a) **Horizontal and vertical components of velocity after time ***t*

Horizontal component of velocity = *v*_{x }= *u Cos q *(this component is not subject to gravity and remains constant throughout the projectile’s flight)

Vertical component of velocity (from (i)):= *v*_{y }= *u Sin q – gt*

(b) Horizontal and vertical components of displacement after time *t*

Horizontal component of displacement = *x *= *v*_{x }*t* = *ut Cos q*

Vertical component of displacement (from (iii)) = *y* = *v*_{y }*t* = *ut Sin q ?gt*^{2}

The equations derived in (a), (b) above are used to derive answers to parts (c) ?(f):

(c) Maximum height reached:

At maximum height, *v*_{y} = 0 Þ *u Sin q – gt = *0

Þ Time elapsed to reach maximum height, *t * =

Þ To calculate maximum height, put this value of *t* into the expression for *y* derived in (b) above: