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Need help with a projectile up an incline problem

  1. Jun 3, 2010 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    for part c

    3. 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.
  2. jcsd
  3. Jun 4, 2010 #2


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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:)
    (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:
  4. Jun 11, 2010 #3
    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.
  5. Jun 12, 2010 #4
    For (c)
    You need to follow tiny tim method. Hello Tiny tim is this correct.
    For vertical axis
    0=(50cos45)t-(0.5 gcos20)t^2
    t=7.525 s
    range=(50cos65)*7.525/cos20 =169.21 m
  6. Jun 12, 2010 #5
    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)


    [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]


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


    [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: Jun 12, 2010
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