If a projectile is fired with an initial velocity of v0...

If a projectile is fired with an initial velocity of $v_{0}$ meters per second at an angle $\alpha$ above the horizontal and air resistance is assumed to be negligible, then its position after $t$ seconds is given by the parametric equations $$x=\left(v_{0} \cos \alpha\right) t \quad y=\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2}$$ where $g$ is the acceleration due to gravity $\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)$ (a) If a gun is fired with $\alpha=30^{\circ}$ and $v_{0}=500 \mathrm{m} / \mathrm{s},$ when will the bullet hit the ground? How far from the gun will it hit the ground? What is the maximum height reached by the bullet? (b) Use a graphing device to check your answers to part (a). Then graph the path of the projectile for several other values of the angle $\alpha$ to see where it hits the ground. Summarize your findings. (c) Show that the path is parabolic by eliminating the parameter.

If a projectile is fired with an initial velocity of $v_{0}$ meters
per second at an angle $\alpha$ above the horizontal and air
resistance is assumed to be negligible, then its position
after $t$ seconds is given by the parametric equations
$$x=\left(v_{0} \cos \alpha\right) t \quad y=\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2}$$
where $g$ is the acceleration due to gravity $\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)$
(a) If a gun is fired with $\alpha=30^{\circ}$ and $v_{0}=500 \mathrm{m} / \mathrm{s},$ when
will the bullet hit the ground? How far from the gun
will it hit the ground? What is the maximum height
reached by the bullet?
(b) Use a graphing device to check your answers to part (a). Then graph the path of the projectile for several
other values of the angle $\alpha$ to see where it hits the
ground. Summarize your findings.
(c) Show that the path is parabolic by eliminating the
parameter.

Key Concepts

Elimination of the Parameter

Eliminating the parameter involves removing time from coupled parametric equations to derive a direct relationship between the horizontal and vertical coordinates. This process typically demonstrates that the projectile's trajectory is a parabola, highlighting the quadratic nature of the vertical displacement as a function of horizontal displacement.

Graphical Verification

Graphical verification uses computational tools or graphing devices to plot the projectile's trajectory. This approach provides a visual check of the analytical results, enabling the observation of key properties like the parabolic shape, range variations with different launch angles, and overall consistency of the modeled motion with theoretical predictions.

Maximum Height

The maximum height is the peak vertical displacement reached by a projectile during its flight. It occurs when the vertical component of the velocity becomes zero before the projectile begins to descend under the influence of gravity, and can be calculated using kinematic equations.

Projectile Motion

Projectile motion describes the behavior of an object that is launched into the air and moves under the influence of gravity, typically assuming that air resistance is negligible. It involves the independent analysis of horizontal and vertical motions, which are influenced separately by the object's initial velocity and gravitational acceleration.

Time of Flight

Time of flight is the total duration that a projectile remains airborne. It is determined by analyzing the vertical component of motion under gravity and solving for the time when the projectile returns to its original vertical position, assuming it starts and finishes at the same height.

Parametric Equations

Parametric equations express the coordinates of a trajectory as functions of an independent variable, usually time. They are particularly useful in projectile motion, where the horizontal and vertical positions can be modeled separately and later combined or transformed, such as eliminating the time parameter to yield the trajectory’s Cartesian form.

Horizontal Range

The horizontal range is the total horizontal distance traveled by the projectile from its launch point to its landing point. It is calculated by multiplying the horizontal component of the initial velocity by the time of flight, and it reflects how far the projectile will travel before hitting the ground.

Transcript

00:02 We're going to consider a projectile which has been launched and whose motion is modeled by these two parametric equations with initial conditions.

00:12 Alpha equals 30 degrees and its initial speed is 500 meters per second.

00:17 So we're first going to find the time at which the projectile hits the ground.

00:23 So if we factor out t in the equation for y, we get t times v0 sine alpha minus, 1 1 half g t so why is 0 either when t is 0 or when t causes this other factor to be 0 we're not interested in when t is 0 that's when we first launch a projectile so we have 0 is equal to v0 sine alpha minus 1 half g t or rearranging t is equal to 2 times v0 sine alpha over g.

01:08 If you plug all this into your calculator, you'll get that that's about 51 .02 seconds.

01:16 Now we want to know how far the projectile is traveled in that time.

01:21 So we'll take this result, 51 .02 seconds.

01:25 Or if you prefer, you can use the original expression for t.

01:29 So x is equal to v0 cosine alpha times 51 .02.

01:37 Seconds and if you plug that into your calculator you'll get it's about 22 ,092 meters okay lastly we want to know at what the maximum height that it achieves is so we'll take the derivative of why and we'll see where that is zero to find a critical point which has to be its maximum value because the two endpoints are both minimums so the derivative of y is equal to v0 sine alpha minus g t.

02:23 So if we want to see where this is zero, 0 is equal to v0 sign alpha minus g t.

02:31 Rearranging t is equal to v0 sine alpha over g.

02:38 So this is the time at which y achieves its maximum.

02:45 If you plug this into your calculator with v0 is equal to 500 and approximating that g is about equal to 9 .8 meters per second squared, then you'll get that this time is about 25 .5 seconds.

03:08 So equals 25 .5.

03:13 Seconds.

03:15 So now let's plug this back into the original equation for y to get the maximum height.

03:20 Y is equal to v0 sine of alpha times 25 .5 seconds minus one half g times 25 .5 seconds squared.

03:39 And again, if you plug that in, then you'll get us about 3 ,100.

03:45 88 .8 meters.

03:50 Okay, with all that calculated, we're going to go look at a graph of these parametric equations now and verify that our results are about right.

04:01 So here we have a graph of our results.

04:04 We're going to go zoom in on this endpoint a bit.

04:09 First off, we can see that the time is right because it's pretty much touching the x -axis.

04:15 It's going to be a little off since it's just an approximation.

04:20 Additionally, we had that it would be at about 22 ,092 meters.

04:27 And look at that.

04:28 It's a little more than 22 ,092 meters.

04:31 So now let's zoom back out and we'll look at the height, the maximum height, which we can see is about here.

04:41 And that happens about halfway through...