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Homework Help: Projectile motion of car driving off a cliff

  1. Aug 6, 2013 #1
    The question is -

    A 1200kg car drives off a cliff with a horizontal velocity of 30m/s and an upward vertical velocity of 15m/s.
    a) Neglecting air resistance, what is its horizontal and vertical velocity after 1s?
    b) What is the maximum height above the ramp the car reaches?

    I used - v = vi + at

    so - vh = 30m/s + 0*1 = 30m/s
    and - vv = 15m/s + 10*1 = 25m/s

    However I am not sure if this is the right formula, and I can't see how to get from question a to question b. If someone could point me in the right direction in regards to formula it would be much appreciated. Thank you :)
  2. jcsd
  3. Aug 6, 2013 #2
    This is correct.

    Why is this PLUS 10*1? What is the direction of acceleration relatively to the direction of initial vertical velocity?

    What can you say about the vertical velocity when the maximum height is reached?
  4. Aug 6, 2013 #3
    Ahh, it should be -10*1 because gravity is causing it to accelerate in the opposite direction? So vv = 5m/s?
    Vertical velocity when maximum height is reached would be 0m/s before beginning to fall..
  5. Aug 6, 2013 #4
    Correct. Always mind the directions and the corresponding signs. They are important.

    Right. Now, does that allow you to compute when this will happen? And knowing when, can you compute the vertical distance?
  6. Aug 6, 2013 #5
    I'm not sure.. I can see there is something there in the car losing 10m/s in 1second but I'm not sure how to calculate how long it would take for it to get to 0m/s.
  7. Aug 6, 2013 #6
    would it be vi + vf/2? so (15m/s + 0m/s)/2 =7.5m?
  8. Aug 6, 2013 #7
    From your equation, ## V_v = V_{vi} - gt ##, at what ## t ## will ## V_v ## be zero?

    No, this cannot be correct. Dimensionally this is velocity, while you need distance.
  9. Aug 6, 2013 #8
    I must of rearranged this formula wrong. I tried to solve for time by -

    vv = vvi - (g*t)
    vv - vvi = -(g*t)
    (vv - vvi) / -g = -t
    T = (vv - vvi)/g

    so t = (0-15) / 10
    so t= -1.5s

    Which isn't a practical solution, as we don't work in negative time!
  10. Aug 6, 2013 #9
    Here is your mistake. You had one minus on the right hand side. Then you moved it to the left hand side, AND retained on the right. That's not correct, because that essentially discards it completely, giving you the wrong sign in the end.
  11. Aug 6, 2013 #10
    Thank you!

    t = (vv - vvi) / -g
    t= (0-15)/-10
    t= 1.5s


    d= 1/2 (vf + vi)*t
    d= 1/2 (0 + 15)*1.5
    d= 0.05m?
  12. Aug 6, 2013 #11
    How can (1/2) * (15) * (1.5) be equal to 0.05?
  13. Aug 6, 2013 #12
    Incorrect use of parentheses! Just testing you :P


    I think I need to go double check all my maths for the rest of my assignment now...
  14. Aug 6, 2013 #13
    Looking good now!
  15. Aug 7, 2013 #14
    The last part of this question is -

    Calculate whether the car will make the the jump if (the distance to the other side) d=55m?

    I don't know how to do this question without knowing the height of the cliff the car is on, and the height that it has to reach on the other side. Does it have something to do with the previous answer of it taking 1.5s to reach a height of 11.25m?
  16. Aug 7, 2013 #15
    I would guess - guess - that the initial and final points of the jump are at an equal height.
  17. Aug 7, 2013 #16
    The problem is I'm not told what the initial and final heights of the jump are, so how can I express this?
  18. Aug 7, 2013 #17
    As I said: I think they should be considered equal. In that case, does it really matter what their heights are?
  19. Aug 7, 2013 #18
    Hmm I understand the concept behind what you are saying, but I don't know how to show that using physics.
  20. Aug 7, 2013 #19
    What does "making the jump" mean?
  21. Aug 7, 2013 #20
    That the car will reach the other side without falling into the abyss?
  22. Aug 7, 2013 #21
    So how can you check that when the car reaches the other side, it has not fallen into the abyss?
  23. Aug 7, 2013 #22
    Well taking into consideration the previous problem where the car reaches 11.25m above the ramp, ie the initial starting point, and if the ramp is the same on the other side - that the car does not drop below 11.25m? That if the car drops below 11.25m then it would miss the ramp?
  24. Aug 7, 2013 #23
    11.25 m is the max height the car reaches ABOVE the cliff. It should not drop BELOW the cliff while it flies over the abyss.
  25. Aug 7, 2013 #24
    Ah sorry I wrote the last part wrong. If it drops lower than 11.25m below its highest point of 11.25m, then it would not make the jump?
  26. Aug 7, 2013 #25
    Well, that is definitely correct, but is it not simpler to say that "not making the jump" is when it drops below the initial height, i.e, zero?

    Anyway, either if these two conditions is correct. So how can you check for those conditions?
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