CK-12-Physics - Intermediate

3.4. Projectile Motion http://www.ck12.org

FIGURE 3.28

Author: Image copyright Konstantin Yol-

shin, 2012; modified by CK-12 Founda-

tion - Christopher Auyeung License: Used

under license form Shutterstock.com

Source: http://www.shutterstock.com

the initialxandycomponents of the velocity are,vcosθandvsinθ, respectively. With this in mind, we rewrite our
kinematic equations as follows:

1.x−direction:xf=R= (vcosθ)t+xi

2.y−direction:yf=^12 gt^2 +(vsinθ)t+yi

The other equations are treated the same way, using the initial velocity in they−direction:

3.vy f=gt+vsinθ

4.vy f^2 = (vsinθ)^2 + 2 g∆y

5.vave=vsinθ 2 +vf y

Special cases

What is the maximum range,R, of a projectile ifyi=yf=0?

setting yiand yf=0 in equation 2, and factoring outt, we have:

t

( 1

2 gt+vsinθ

)

= 0

There are two solutions:

t=0 andt=−^2 vgsinθ

The trivial condition is satisfied at launch(t= 0 ,yi= 0 )and the nontrivial condition is satisfied at landing

(

t=−2 sing θ,yf= 0

)

.

Ift=−^2 vgsinθis substituted into equation 1 :R= (vcosθ)t

We have:R= (vcosθ)

(

− 2 vsinθ

g

)

=−^2 v

(^2) sinθcosθ
g
Using the trigonometric identity 2 sinθcosθ=sin 2θ, we have:
R=−v
(^2) sin 2θ
g
We can extract a useful result from the range equation. Assuming thatgis constant, the range is a function of theV
andθ. Let’s consider the angle. Since the maximum value of the sine is 1.0 then whatever angle makes sin 2θ=1,
will also maximize the range. Since sin 90◦=1, 2θ= 90 ◦and the angle which produces the greatest range is 45◦.
The range continually increases with increasing velocity so there is no “interesting” information we can get out of