Projectile Motion distance problem

[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 – .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.

See attachment.

2. The attempt at a solution

(a) I let y=0 since it will strike the ground at this point on the profile. Then I solved for x:

0 = 0.4x – .006x²
0= x(0.4 – 0.006x)
X=0 & x=66.667m
So my coordinates for the projectile when it strikes the ground profile will be: (66.67, 0)

(b) I used the following:

x=66.667
x = x_o + (v_o)yt
t = 4.35 seconds

v_y = (v_oy) - at
v_y = -29.81 m/s

v_x = (v_o)x = 15.32 m/s

v = √ (-29.81)² + (15.32)² = 33.51 m/s


c) I then used the equation:
v² = (v_o)y² + 2(-9.81)h
h = 8.43 m

d) Assuming that all of the above is correct, I'm not sure how to find this part. I initially thought that it was the same as part c.

 

[iRaid]

For c you can use t/2 to get the max height (since the top of the parabola would be the highest point) and I don't understand d, it seems like the same as c? Maybe they want the x coordinate for one of them and the y coordinate for the other.

 

[aaronfue]

For c you can use t/2 to get the max height (since the top of the parabola would be the highest point) and I don't understand d, it seems like the same as c? Maybe they want the x coordinate for one of them and the y coordinate for the other.

Sorry, I was editing my question. I did find part c, but part d is still not clear to me.

I'd appreciate it if you can check my work for parts a and b?

 

[iRaid]

Well for c I don't get the same answer, but I might be wrong.
y=(20sin40)t-4.9t^2
t/2=2.175 (plug into the above)
y=4.781m

 

[aaronfue]

Well for c I don't get the same answer, but I might be wrong.
y=(20sin40)t-4.9t^2
t/2=2.175 (plug into the above)
y=4.781m

Here is what I did to get h:

v² = (v_o)y² + 2(-9.81)h
0 = (20sin40°)² - 19.62h
0 = 12.86² - 19.62h
0 = 165.38 - 19.62h
19.62h = 165.38
h = 8.429 m = 8.43 m

Another thing that I was just thinking about was that the total horizontal distance would not be, by regular calculations, 66.67 m. It came out to be 40.14 m. But the 66.67m is along the curve y=0.4x - 0.006x². And that is something I'm not sure how to calculate. Any ideas?

 

[haruspex]

(a) I let y=0 since it will strike the ground at this point on the profile.

That's wrong. It starts at (0, 0). Where it strikes the ground y may be different.

 

[aaronfue]

That's wrong. It starts at (0, 0). Where it strikes the ground y may be different.

How would I find the correct distance along that path?

 

[haruspex]

Can you find an equation for the trajectory (y as a function of x)?

 

[aaronfue]

Can you find an equation for the trajectory (y as a function of x)?

I've graphed the projectile and I believe I have the equation that describes the trajectory.

y = -0.021x² + 0.839x

By graphing (using excel), I also found the coordinates of where the projectile impacts the ground (represented by: y = 0.4x - 0.006x²): x=29.47, y=6.58

 

[haruspex]

I've graphed the projectile and I believe I have the equation that describes the trajectory.

y = -0.021x² + 0.839x

Looks right, but did you extract that by fitting to the graph or did you obtain it by algebra? Can you write it as a function of unknown launch speed and angle, u and theta?

By graphing (using excel), I also found the coordinates of where the projectile impacts the ground (represented by: y = 0.4x - 0.006x²): x=29.47, y=6.58

OK, but again you should be able to do this by algebra. It's quite easy.