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Homework Help: Kinematic problems with angles

  1. Oct 18, 2012 #1
    This is my first year taking a Physics class, so this might just be a simple question but I'm stuck.

    1. The problem statement, all variables and given/known data

    A stunt driver wants to make his car jump over eight cars parked side by side below a
    horizontal ramp. (a) With what minimum speed must he drive off the horizontal ramp?
    The vertical height of the ramp is 1.5m above the cars and the horizontal distance he must
    clear is 20m. (b) If the ramp is no tilted upward, so that the “take off” angle is 10° above
    the horizontal, what is the new minimum speed?

    Note: I already solved part A and got an answer of 35.71m/s

    2. Relevant equations

    Velocity = Initial Velocity + (Acceleration x Time)
    Distance = Initial Distance + (Initial Velocity xTime) + 1/2(Acceleration)(Time ^2)
    Velocity ^2 = Initial Velocity ^2 + 2acceleration(Distance - Initial Distance)

    There might be others too I don't really know, sorry.

    3. The attempt at a solution

    As I said before I already got the answer to part A. Did that by simply finding the time, then solving for velocity. What's throwing me off now is the addition of the 10 degree incline. If someone could maybe show me how to account for that slope of the take off ramp, that would be awesome.
  2. jcsd
  3. Oct 18, 2012 #2


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    Welcome to PF,

    What's the horizontal component of the velocity at launch for the second case? Does it change during flight?

    What's the vertical component? This will tell you the max height reached, and therefore the time spent in the air.
  4. Oct 18, 2012 #3
    I think, not sure, that the vertical velocity is -5.488m/s, and the horizontal one is 35.71m/s.

    Edit: I think those could be wrong, since there's probably going to be a new time with the 10 degree angle.
    Last edited: Oct 18, 2012
  5. Oct 18, 2012 #4


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    Yeah, I mean, the time is going to be longer, because instead of just free-falling from some starting height, it first has to go up to some maximum height, and then free fall this larger distance. You can work out how high it gets in a straightforward way though, if you know the initial vertical speed.
  6. Oct 18, 2012 #5
    Alright, so what I just did was find the real initial velocities (X = 19.7m/s, Y = 3.47m/s)
    Then with that and also assuming the ending velocity is 0 (since it will be at the max height it will not be moving up or down) I did Vy = Viy + (a)(t) or with my numbers, 0 = 3.47 + (-9.8)(t) and got t = .354s. This would be how long it takes to get to the max height.
    Then with that information I calculated the Max Height to be somewhere around .615m.

    Not sure where to go from here? I think I would have to solve for Vx to get my answer but I'm not sure.

    Just calculated Vx, and I got the same thing as Vix (19.7m/s) which would make sense I guess according to the equations I have. Not too sure though
  7. Oct 19, 2012 #6
    How did you get those velocities without solving the rest of the problem?
  8. Oct 19, 2012 #7
    I don't know really, I was hoping someone could help me with it because I was 99% sure I was wrong.
  9. Oct 19, 2012 #8


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    Yeah, I had things a little bit backwards in my advice to you. I thought you were given the launch speed and angle, and were supposed to figure out how far it would get. But you're given the launch angle and how far it gets, and you're supposed to figure out the launch speed. No matter. All the physics is the same, all that's changed is which variables are unknown and which ones are given.

    So, you have an initial launch speed v, which you should treat as the unknown that you are trying to solve for. It's a vector pointing 10 degrees above the horizontal. Using this information, you can resolve this vector into x and y components (it's just basic trig.). The components v_x and v_y are *also* unknowns, but you can use the angle (and trig) to express them *in terms of v*. That's step 1.

    As for v_x: yes of COURSE it is unchanged throughout the flight. Is there any acceleration in the x direction? No. So why would the x velocity change?

    Step 2: since the motion in x is just motion at a *constant speed*, it's really straightforward to figure out how much time is needed (at that speed) to go a horizontal distance of 20 m. Hint: when you're just moving at a constant speed, how are distance, speed, and time all related to each other?

    Step 3: So now you have a mathematical expression for the flight time t that is required, *in terms of* v_x and the given distance, d. This means that it's actually expressed in terms of your unknown v, the angle theta, and the distance d. (Keep all this algebraic and don't plug in numbers until the end -- it just makes things cleaner and easier). So now step three is to figure how what t is *in terms of v_y* Note that t is the entire flight time, which means it is the time required to rise from 1.5 m up to the max height, PLUS the time required to fall back down to the ground from the max height. This step is definitely a bit tricky and involves multiple steps, because first you'll have to compute the time t1 required to reach the max height (*in terms of* the unknown initial launch speed v_y), and then you'll have to calculate what that max height is (in terms of the unknowns), and then you'll have to calculate the time t2 required to fall to the ground from that max height, and then t = t1 + t2.

    Step 4: So now you have two mathematical expressions for t, one in terms of v_x, that you determined based on the horizontal distance that needs to be travelled, and the other in terms of v_y, that you calculated based on how v_y determines the time spend in the air. These two expressions have to be equal to each other, because they're both equal to t. So, equating them, you should get an expression that has only your unknown v, and other *known* quantities (like the launch angle, the horiztonal distance travelled, and the acceleration due to gravity). Solve for v.
  10. Oct 19, 2012 #9
    In step 3, you don't need to go through the many steps of calculating the max height and time, then fall time. You have a quadratic equation for the height as a function of time, so you can use the quadratic formula to find t in one step.

    On the other hand, you've already got t as function of v, d, and θ. You can plug that in to the equation for projectile motion and you'll get a function that only depends on the (inverse) square of the velocity. Solving for v should be pretty easy from there.
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