MTH 1 - Project II

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 principal 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 feet per second in a direction 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 that the velocity in the *x*-direction (*horizontal*) is feet per second. So, the distance traveled horizontally at the end of *t* seconds is feet.

Also, the projectile starts with a *vertical* component of velocity of feet per second. This velocity would cause the projectile to rise upward to a height of feet in *t* seconds.

These equations for the path and speed are called parametric equations because they depend on a parameter *t*.

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.

Our picture doesn’t take into account the pull of gravity, which lessens the distance traveled. According to a formula of physics, the amount to be subtracted is from the vertical position is , where *g* is a constant approximately equal to 32.

Write the parametric equations for the path, for *x* and *y* in terms of *t*:

x = v_x*t (v_x = velocity component)

y = h_0-v_y*t-.5*32*t^2

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 the shape of a parabola.

We will use the parametric equations you found in 2) to solve the following problem: *A stone is thrown with a velocity of 160 ft./sec in a direction 45° above the horizontal. Find how far away the stone strikes the ground and its greatest height.*

Substitute , α = 45°, and g = 32 into your equations in 2). Write the resulting equations below:

The stone reaches the ground when *y* = 0. Substitute this value for *y* into the second equation and find *t*, the time of flight:

Use the* t*-value you found in b) in the first equation to find *x*, how far away the stone strikes the ground:

We know that the stone moves along a parabola that opens downward and that a parabola is symmetric with respect to its axis. Hence the greatest height is the value of y when *t* is half the flight time.

Find *t/2* using the *t*-value from b), and substitute it into the second equation to find y, the maximum height:

We will now use the rectangular form of the equation for the path of the projectile, that you found in 3), to solve the same problem.

Substitute , α = 45°, and g = 32 into your equation from 3). Write the resulting equation below:

The stone reaches the ground when *y* = 0. Substitute this value for *y* into the equation and find (*x*, 0), the point where the stone strikes the ground.

The vertex is the highest point of the parabola. Use the formula to find the vertex of , the point of maximum height of the projectile.

Subject | Mathematics |

Due By (Pacific Time) | 10/12/2015 12:00 am |

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