Projectile - 2D motion - incline - acceleration

[KUphysstudent]

Homework Statement

A test rocket is fired from rest at point (A) along (b) a 200m incline, accelerating 1.25 m/s²

Given:
b = 200 m, the angle at point A is 35°, a = 1.25 m/s²

a) Find the maximum height above the ground.
b) Find the greatest range.

Homework Equations

If it was a constant velocity I would have no problem solving this but I don't know how to deal with this acceleration. My book doesn't seem to help me, it is rated the most difficult in the chapter of 2-3D motions. but there are no examples with varing acceleration in the horizontal direction.

When velocity is given
v_0y = v₀ sinθ
v_y = v_0y - gt₁ = 0
t₁ = v_oy/g
h = v_oyt₁ - ½gt₁²

v_ox = v₀ cosθ
y = 0 = v_oyt₂ - ½gt₂² = t₂(v_oy - ½gt₂)
t₂ = 2v_oy/g
R = v_oxt₂

But how do I find these when its acceleration?

The Attempt at a Solution

 

[Geofleur]

Hint: Consider dividing the problem into two parts.

 

[KUphysstudent]

Hint: Consider dividing the problem into two parts.

does it make sense to find the final velocity at the 200 m mark on the ramp then calculate the height of the triangle and then go from there?

 

[BvU]

Yes. That is splitting in two parts...
And: you seem to think the horizontal acceleration is varying. Why ?

 

[KUphysstudent]

Yes. That is splitting in two parts...
And: you seem to think the horizontal acceleration is varying. Why ?

Now that you say that it is very clear that it is! thank you so much <3

 

[BvU]

So, just for the sake of other readers ( ), where are we now ?

 

[KUphysstudent]

So, just for the sake of other readers ( ), where are we now ?

I went to bed, but today we're at this point:

v_f² = v_i² + 2 * a * d
v_f = sqrt(0 + 2 * 1.25 m/s² * 200 m) = 22.361 m/s

The height of the triangle is, sin(35°)*200m = 114,7m = a

-114.7m = (22.361 m/s * sin(35°))*t+(-½*9.8 m/s² * t²)
0 = -½*9.8m/s²*t²+sin(35°)*22.361m/s*t+114.7m
Δt = (-sin(35°)*22.361m/s ±sqrt((sin(35°)*22.361m/s)-4*(-1/2*9.8)*114,7m)) / -9.8 m/s²
Δt =+√ = -5,73s , Δt = -√ = 3.776s

v_0y = v₀ sinθ
v_y = v_oy - gt₁ = 0
t₁ = v_0y/g
h = v_0yt₁ - ½gt₁²

v_0y = sin(35°)*22,361 m/s = 12,83 m/s
v_y = 12,83 m/s - 9.8m/s²*t₁ = 0
t₁ = v_oy/g = 12.83 m/s/9.8m/s² = 1.3s
h = v_0yt₁ - ½gt₁² = 12,83m/s*1.3s-½*9.8m/s²*1.3²₂ = 8,398m
114,7m + 8.398m = 123,098m
This doesn't make sense to me, 8 meters height from a rocket, I know there are different rockets I just expected more. (its a huge ramp for 8 meters of height)

v_ox = 22.361 m/s * cos(35°) = 18,3 m/s
y = 0 = v_oy * t₂ - ½ * 9.8 m/s² *t² = t₂(v_oy - ½ * g * t₂)
t₂ = (2*v_oy)/9.8 m/s² = 2,62s
R = v_ox*t₂ = 18,3 m/s * 2,62s = 47,96m
but that length is still 114,7m above the ground.

x = (v₀*cos(-35°)*t = 22,361 m/s * cos(-35°)*3,776s = 67,17m
47,96m + 67,17m = 117,125m

Ok I am pretty confused at this point and pretty sure I made a lot of mistakes, unfortunately.

 

[BvU]

Oh boy, what a long post. And then I have to ask where the first line comes from (I don't see it in the relevant equations, that's why I ask not the I went to bed, but the v_f² line of course)

 

[KUphysstudent]

Oh boy, what a long post. And then I have to ask where the first line comes from (I don't see it in the relevant equations, that's why I ask not the I went to bed, but the v_f² line of course)

They come from this site http://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations
They're calling these equations the big four:

 

[BvU]

Good. I've had some coffee since. Big four sounds impressive and is right here. Let me continue... a bit slower

Note that the d in the first one of these big four first is distance covered (*) so for your h that meant height counting from leaving the ramp...

[edit] more coffee ⇒ yeah, 8 m counting from the top of the ramp. It's not a fast rocket, I admit...
more like a radio-controlled toy car.
All because 1.25 m/s² isn't all that much.​

(*) see $x = x_0 + v_0 t + (1/2) a t^2\$ here in our own big toolkit.

 

[KUphysstudent]

Good. I've had some coffee since. Big four sounds impressive and is right here. Let me continue... a bit slower

Note that the d in the first one of these big four first is distance covered (*) so for your h that meant height counting from leaving the ramp...

[edit] more coffee ⇒ yeah, 8 m counting from the top of the ramp. It's not a fast rocket, I admit...
more like a radio-controlled toy car.
All because 1.25 m/s² isn't all that much.​

(*) see $x = x_0 + v_0 t + (1/2) a t^2\$ here in our own big toolkit.

that is so sad with the d :(

 

[KUphysstudent]

Good. I've had some coffee since. Big four sounds impressive and is right here. Let me continue... a bit slower

Note that the d in the first one of these big four first is distance covered (*) so for your h that meant height counting from leaving the ramp...

[edit] more coffee ⇒ yeah, 8 m counting from the top of the ramp. It's not a fast rocket, I admit...
more like a radio-controlled toy car.
All because 1.25 m/s² isn't all that much.​

(*) see $x = x_0 + v_0 t + (1/2) a t^2\$ here in our own big toolkit.

also I am a bit confused with this equation how would I find the initial velocity at the end/top of the ramp.
all I know is x = 200 + 0 + 1/2 * 1.25 * t^2, I don't know either x or t

 

[BvU]

At the end of the ramp you already found v = 22.36 m/s (v_x = 18.32 m/s, v_y = 12.82 m/s). From there you also found a flight time of 3.77 sec (I found 3.704, actually (using g = 9.8 m/s²) -- perhaps you can check; the other one was -6.32 s) ).

From that point the horizontal motion is uniform, so distance covered is just $v_{0,x} \, t$, your 67.2 m (my result was 67.8).

I think you did well, but made it more complicated than necessary...

 

[KUphysstudent]

At the end of the ramp you already found v = 22.36 m/s (v_x = 18.32 m/s, v_y = 12.82 m/s). From there you also found a flight time of 3.77 sec (I found 3.704, actually (using g = 9.8 m/s²) -- perhaps you can check; the other one was -6.32 s) ).

From that point the horizontal motion is uniform, so distance covered is just $v_{0,x} \, t$, your 67.2 m (my result was 67.8).

I think you did well, but made it more complicated than necessary...

I used v = 22,361 m/s v_x = 18,3 m/s v_y = 18,3 m/s flight time = 3,776s using g = 9,8 m/s² the negative time had a larger difference compared to yours, I have -5,73s

 

[KUphysstudent]

I think I made a mistake during the flight time calculation

(-sin(35)*22.361-sqrt((sin(35)*22.361)-4(-1/2*9.8)*114.7))/-9.8 = 6.16s
(-sin(35)*22.361+sqrt((sin(35)*22.361)-4(-1/2*9.8)*114.7))/-9.8 = -3,54s

This is using a calculator I know how works, the other numbers was from Maple which i think always calculate in radians unless you do the 1/180*pi conversion

 

[BvU]

Excel (I didn't do any rounding - except g = 9.8) says
(PF won't use courier for code ? )

10 h0 114.7153 =200*SIN(35/180*PI())
11
12
13 v0^2 500 =2*1.25^1*200
14
15 v0 22.36068 =SQRT(D13)
16
17 v0x 18.3168 =D15*COS(35/180*PI())
18 v0y 12.82556 =D15*SIN(35/180*PI())
19
20 t gnd 3.70366 =(0-D18+SQRT(D18^2+4*D10*9.8/2))/9.8
21
22 v0x * t gnd 67.83919 =D20*D17

 

[KUphysstudent]

Excel (I didn't do any rounding - except g = 9.8) says
(PF won't use courier for code ? )

10 h0 114.7153 =200*SIN(35/180*PI())
11
12
13 v0^2 500 =2*1.25^1*200
14
15 v0 22.36068 =SQRT(D13)
16
17 v0x 18.3168 =D15*COS(35/180*PI())
18 v0y 12.82556 =D15*SIN(35/180*PI())
19
20 t gnd 3.70366 =(0-D18+SQRT(D18^2+4*D10*9.8/2))/9.8
21
22 v0x * t gnd 67.83919 =D20*D17

Yep I forgot to square b under the root.

so my delta t's er now -3,7s and 6,32s

Btw don't you have to divide by negative g and not positive?

 

[BvU]

Btw don't you have to divide by negative g and not positive?

Oops. Brilliant ! Sounds very reasonable to me and gets the projectile a little further as well. Perhaps we should top it off with a drawing showing the trajectory (y versus x) !

[edit] Must have more coffee (again): at the launch from the top of the ramp back in time the parabola should be shorter than forward, so physical intuition should have kicked in and objected right away !

 

[KUphysstudent]

Oops. Brilliant ! Sounds very reasonable to me and gets the projectile a little further as well. Perhaps we should top it off with a drawing showing the trajectory (y versus x) !

[edit] Must have more coffee (again): at the launch from the top of the ramp back in time the parabola should be shorter than forward, so physical intuition should have kicked in and objected right away !

As you can see from my earlier paint drawing, drawing is not a set of skills I was blessed with.

And I guess the questions are done if I just fix the b^2 and correct the calculations using the time.

 

[KUphysstudent]

v_f² = v_i² + 2 * a * d
v_f = sqrt(0 + 2 * 1.25 m/s² * 200 m) = 22.361 m/s

The height of the triangle is, sin(35°)*200m = 114,7m = a

-114.7m = (22.361 m/s * sin(35°))*t+(-½*9.8 m/s² * t₂)
0 = -½*9.8m/s²*t₂+sin(35°)*22.361m/s*t+114.7m
Δt = (-sin(35°)*22.361m/s ±sqrt((sin(35°)*22.361m/s)²-4*(-1/2*9.8)*114,7m)) / -9.8 m/s²
Δt =+√ = -3,7s , Δt = -√ = 6,32s

v_0y = v₀ sinθ
v_y = v_oy - gt₁ = 0
t₁ = v_0y/g
h = v_0yt₁ - ½gt₁²

v_oy = sin(35°)*22,361 m/s = 12,83 m/s
v_y = 12,83 m/s - 9.8m/s²*t₁ = 0
t₁ = v_oy/g = 12.83 m/s/9.8m/s² = 1.3s
h = v_0yt₁ - ½gt₁² = 12,83m/s*1.3s-½*9.8m/s²*1.322 = 8,398m
114,7m + 8.398m = 123,098m
This doesn't make sense to me, 8 meters height from a rocket, I know there are different rockets I just expected more. (its a huge ramp for 8 meters of height)

v_ox = 22.361 m/s * cos(35°) = 18,3 m/s
y = 0 = v_oy * t₂ - ½ * 9.8 m/s² *t² = t₂(voy - ½ * g * t₂)
t₂ = (2*v_oy)/9.8 m/s² = 2,62s
R = v_ox*t₂ = 18,3 m/s * 2,62s = 47,96m
but that length is still 114,7m above the ground.

x = (v₀*cos(-35°)*t = 22,361 m/s * cos(-35°)*6,32s = 115,76m
47,96m + 115,76m = 163,7m

Something tells me this last part from x = ... is incorrect.
Is 22,361 m/s * cos(-35°)*6,32s the range of the rocket from the top of the ramp and to the ground 114,7m below?