Projectile Motion highest point

[aaronfue]

Homework Statement

At t = 0, a projectile is located at the origin and has a velocity of 20 m/s at 40° above the horizontal. The profile of the ground surface it strikes can be approximated by the equation y = 0.4x – 0.006x², where x and y are in meters. Determine:

(a) the approximate coordinates of the point where it hits the ground,
(b) its velocity and direction when it hits the ground,
(c) its highest point in flight and,
(d) the greatest distance above the ground.

*****Excel table and graph are attached.

2. The attempt at a solution

a) Determine the approx. coordinates of the point.
I plotted the data in excel and obtained the equations for both. I tried to figure this out algebraically, but my answers were way off. My values for x were 0 and 12.95. So I plugged the two equations in my Ti-89 titanium and checked the intersection with that. And it made more sense.
x = 29.47 m
y = 6.58 m

So I believe that this is my answer for part a.

b) Velocity and direction when it hits the ground?
I’m a bit stuck on figuring this part out. Any assistance on this would be great! I think I can use the x and y coordinates from part a to figure this out? Maybe?

c) Highest point in flight?
I think this meant without considering the equation of the surface. So, I used the following equation: v² = (v_o)_y² + 2*a*h
h = 8.43 m

d) Greatest distance above the ground?
I believe this means to consider the surface equation and find the greatest distance. This part I’m also having trouble with. I’m sure it’s something I have done, but I can’t remember.

 

[voko]

The equation of motion that you have found is correct. That is very good. You can use it to solve all the parts.

The results you obtained for part a numerically are slightly off. It is not clear why you could not solve the part algebraically, perhaps you need to show your attempt.

For part b, knowing the coordinate x of the impact point, you should be able to determine the time of impact, and from that, the vertical velocity component.

For c, you answer is close. Perhaps you need to show how you got it.

For d, you need to compose the equation of the distance from the ground at any coordinate x. Then find its maximum.

 

[aaronfue]

The equation of motion that you have found is correct. That is very good. You can use it to solve all the parts.

The results you obtained for part a numerically are slightly off. It is not clear why you could not solve the part algebraically, perhaps you need to show your attempt.

For part b, knowing the coordinate x of the impact point, you should be able to determine the time of impact, and from that, the vertical velocity component.

For c, you answer is close. Perhaps you need to show how you got it.

For d, you need to compose the equation of the distance from the ground at any coordinate x. Then find its maximum.

Thanks for checking my work.

Part A:
After I graphed the data and obtained my equation for the path of the projectile, I did what I remembered and set each equation equal to each other and solved for x, then substituted that into one of the equations and solved for y.

My answers were: x=0/x=21.95 & y=6.75
Now my y coordinate seems to be okay, but the x=21.95 seems off. And when I checked the excel graph, that number seemed like the vertex of the path.

Part B:
Since I used my original impact coordinate I was able to find the time until it reached the impact coordinates (29.47m, 6.58m).
x = x_o + (v_o)_xt
t = 1.9235 seconds
Then use that to find the final y velocity.
v_y = (v_o)_y +at
v_y = 6.01 m/s

Part C:
I used the following equation:
v² = (v_o)_y² + 2a(h)
0 = (20sin40°)² - 2*(9.81)*h
h = 8.43 m

I'm not sure I follow what you mean for part D.

 

[voko]

Part A:
After I graphed the data and obtained my equation for the path of the projectile, I did what I remembered and set each equation equal to each other and solved for x, then substituted that into one of the equations and solved for y.

My answers were: x=0/x=21.95 & y=6.75

You have the correct equation of motion $y = -0.021 x^2 + 0.839 x$. As you say, you should equate this with $y = -0.006 x^2 + 0.4 x$, getting $-0.021 x^2 + 0.839 x = -0.006 x^2 + 0.4 x$, or, simplifying, $(-0.015 x + 0.439) x = 0$. It is satisfied when $x = 0$, which we don't care about, and when $x = 0.439/0.015 = 29.267$.

How could you get x = 21.95?

Part B:
Since I used my original impact coordinate I was able to find the time until it reached the impact coordinates (29.47m, 6.58m).
x = x_o + (v_o)_xt
t = 1.9235 seconds
Then use that to find the final y velocity.
v_y = (v_o)_y +at
v_y = 6.01 m/s

Except that you should use the correct value of x, your steps are correct. You should just continue: since you know both the horizontal and vertical components of velocity, you should be able to find its magnitude and angle.

I'm not sure I follow what you mean for part D.

What is the distance above the ground at some coordinate x?

 

[aaronfue]

You have the correct equation of motion $y = -0.021 x^2 + 0.839 x$. As you say, you should equate this with $y = -0.006 x^2 + 0.4 x$, getting $-0.021 x^2 + 0.839 x = -0.006 x^2 + 0.4 x$, or, simplifying, $(-0.015 x + 0.439) x = 0$. It is satisfied when $x = 0$, which we don't care about, and when $x = 0.439/0.015 = 29.267$.

How could you get x = 21.95?



Except that you should use the correct value of x, your steps are correct. You should just continue: since you know both the horizontal and vertical components of velocity, you should be able to find its magnitude and angle.



What is the distance above the ground at some coordinate x?

I made a calculation error. I get the same answer as you do now.

 

[aaronfue]

You have the correct equation of motion $y = -0.021 x^2 + 0.839 x$. As you say, you should equate this with $y = -0.006 x^2 + 0.4 x$, getting $-0.021 x^2 + 0.839 x = -0.006 x^2 + 0.4 x$, or, simplifying, $(-0.015 x + 0.439) x = 0$. It is satisfied when $x = 0$, which we don't care about, and when $x = 0.439/0.015 = 29.267$.

How could you get x = 21.95?



Except that you should use the correct value of x, your steps are correct. You should just continue: since you know both the horizontal and vertical components of velocity, you should be able to find its magnitude and angle.



What is the distance above the ground at some coordinate x?

For part d, would I have to find the difference of the two parabolas, differentiate w/respect to x, solve for x and then plug that into solve for y?

 

[voko]

Yes, your program is correct.