Need help with a projectile up an incline problem

  • Thread starter daniel1919
  • Start date
  • #1

Homework Statement


A projectile is launched at an angle 45degrees with respect to the horizontal and up a hill that is inclined 20degrees to the horizontal. The initial speed of the projectile is 50m/s. Neglect air resistance.
a)what is maximum height? - I found 63.8m which is correct
b) what is the speed of projectile 2.5s after launching? - 37m/s which is correct
c) find the distance up the hill the projectile lands.
d)calculate the magnitude and direction of the velocity of the projectile at impact.



Homework Equations


for part c
x=vicos(45)t-.5gsin(25)t^2
x=visin(45)t-.5gcos(25)t^2


The Attempt at a Solution


The only idea i have for part c is to change the x-y axis so that x is along the slope. Any help solving or ideas would be great.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi daniel1919! Welcome to PF! :smile:

(have a degree: º and try using the X2 and X2 tags just above the Reply box :wink:)
c) find the distance up the hill the projectile lands.

x=vicos(45)t-.5gsin(25)t^2
x=visin(45)t-.5gcos(25)t^2

(i assume one of those is y ? :rolleyes:)

ok, you have the equations for axes along and perpendicular to the slope …

now put y = 0 to get t, and then substitute into x …

what do you get? :smile:
 
  • #3
754
1
The projectile takes a parabolic path. What is the equation of that path (forget about the hill, for now)?

The hill describes a line; what is it's formula?

Assume the origin to be the point of launch which lies on both the parabola and the hill.
Find the other intersection.
 
  • #4
99
1

Homework Statement


A projectile is launched at an angle 45degrees with respect to the horizontal and up a hill that is inclined 20degrees to the horizontal. The initial speed of the projectile is 50m/s. Neglect air resistance.
a)what is maximum height? - I found 63.8m which is correct
b) what is the speed of projectile 2.5s after launching? - 37m/s which is correct
c) find the distance up the hill the projectile lands.
d)calculate the magnitude and direction of the velocity of the projectile at impact.



Homework Equations


for part c
x=vicos(45)t-.5gsin(25)t^2
x=visin(45)t-.5gcos(25)t^2


The Attempt at a Solution


The only idea i have for part c is to change the x-y axis so that x is along the slope. Any help solving or ideas would be great.

For (c)
You need to follow tiny tim method. Hello Tiny tim is this correct.
For vertical axis
y=0
0=(50cos45)t-(0.5 gcos20)t^2
t=7.525 s
range=(50cos65)*7.525/cos20 =169.21 m
 
  • #5
754
1
That's not what I came up with...

First of all we have:

[tex]V_i = 50 m/s[/tex] (initial velocity)
[tex]\alpha = 45^\circ[/tex] (angle of trajectory)
[tex]\beta = 20^\circ[/tex] (angle of the hill)


Therefore,

[tex]V_{ix} = V_i \times cos(\alpha) = V_i \times cos(45^\circ)[/tex] (initial horizontal velocity)
[tex]V_{iy} = V_i \times sin(\alpha) = V_i \times sin(45^\circ)[/tex] (initial vertical velocity)


but since [itex]cos(45^\circ) = sin(45^\circ)[/tex], we have

[tex]V_{ix} = V_{iy}[/tex]


We know that

[tex]t = \frac{X}{V_{ix}}[/tex]

and

[tex]Y = V_{iy} \cdot t - 4.9t^2[/tex]

so

[tex]Y = V_{iy}\cdot \frac{X}{V_{ix}} - 4.9\left(\frac{X^2}{V_{ix}^2}\right)[/tex]

but since [itex]V_{iy} = V_{ix}[/tex], we have

[tex]Y = X - 4.9 \cdot \left( \frac{X^2}{V_{ix}^2} \right)[/tex] (equation A)


The slope of the hill is [itex]m = tan(\beta) = tan(20^\circ)[/tex] so the equation for the hill is

[tex]Y = X \cdot tan(20^\circ)[/tex]


Substituting for Y in the parabola equation (equation A) gives us

[tex]X \cdot tan(20^\circ) = X - 4.9 \cdot \left( \frac{X^2}{V_{ix}^2} \right) [/tex]

[tex]tan(20^\circ) = 1 - 4.9 \cdot \left( \frac{X}{V_{ix}^2} \right) [/tex]

[tex]\left(\frac{4.9}{V_{ix}^2}\right) \cdot X = 1 - tan(20^\circ)[/tex]

[tex]X = \frac{V_{ix}^2(1 - tan(20^\circ))}{4.9} = \frac{(50 \cdot cos(45^\circ))^2 \cdot (1 - tan(20^\circ))}{4.9}[/tex]

[tex]X = \frac{35.355^2 \cdot 0.636}{4.9} = \frac{1250 \cdot 0.636}{4.9} = \frac{795.037}{4.9} = 162.252[/tex]


Plug this value back into equation A, above, and you get Y = 59.055

The distance up the his is found using the Pythagorean Theorem:

[tex] D^2 = X^2 + Y^2 = 162.252^2 + 59.055^2 = 29813.373[/tex]

and D = 172.665 m

(note: I only rounded off in the text, not in my actual calculations)
 
Last edited:

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