Equations of Projectile Motion

MTH -1
Project - II
<STUDENT NAME>
APRIL 17, 2019

The equations of some curves can be determined more readily by the use of a parameter
than otherwise. In fact, this is one of the principals uses of parametric equations. We
will use parametric equations to help us determine the path of a projectile through the
air.
Suppose that a body is given an initial upward velocity of u feet per second which makes
an angle α with the horizontal. If the resistance of the air is small and can be neglected
without great error, the object will move subject to the force of gravity. This means that
there is no horizontal force to change the speed in the horizontal direction.
If we take the origin to be the point at which the projectile is fired, we see the velocity in
the x-direction (horizontal) is u feet per second. So the distance travelled horizontally at
the end of t seconds is ut feet.
Also, the projectile starts with a vertical component of velocity of v feet per second. This
velocity would cause the projectile to rise upward to a height of h feet in t seconds.
These equations for the path and speed are called parametric equations because they
depend on a parameter t.
a) Draw a picture of the path of the projectile, starting at the origin. Label the x
and y components of the path at some time t>0.
Sol.
-1 0 1 2 3 4 5
-1
0
1
2
3
4
5
b) Our picture doesn’t take into account the pull of gravity, which lessens the
distance travelled. According to a formula of physics, the amount to be
subtracted is from the vertical position is, where g is a constant approximately
X component
y component
u ft/sec

equal to 32. Write the parametric equations for the path, for x and y in terms of
t.
Sol.
-1 0 1 2 3 4 5
-1
0
1
2
3
4
5
x=u cos α t ... ... ...( 1)
v y=u sin α −¿
y= ((u sin α )t− 1
2 g t 2
)... ... ...(2)
c) If we solve the first equation for t and substitute the result in the second, we
obtain the equation of the path in rectangular form. Do this to find the
rectangular form for the equation of the path. The above equation should be of
second degree in x, and first degree in y. Thus, it represents a path in a shape of
a parabola.
Sol.
x=u cos α t ... ... ...( 1)
t = x
u cos α
y= ((u sin α )t− 1
2 g t 2
)... ... ...(2)
y=u sin α ( x
u cos α )− 1
2 g ( x
u cos α )
2
y=x tan ∝− g
2 u2 (cos α )2 x2
g ft/sq. sec
u ft/sec