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Homework Help: Projectile Motion, leaping the river

  1. Oct 27, 2007 #1
    i'm curious to know whether my method is the long way or perhaps, the only way in solving for initial velocity for this problem

    if you have a faster way, i'd like to see it ... i'm trying to be efficient!

    The take off ramp is inclined at 53.0, the river is 40.0m wide, and the far bank is 15.0m lower than the top of the ramp. The river itself is 100m below the ramp. (Ignore air resistance) a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank?


    http://www.mathlinks.ro/Forum/latexrender/pictures/e/b/2/eb21343a1e34a55256a7b86eed1fdf89c8f1b0f8.gif [Broken]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 28, 2007 #2


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    In this problem y and g are in the down ward direction and initial velocity is in the upward direction. Therefore y and g must have the same sign and velocity should be opposite sign.
    There is no shortcut method to solve this problem.
    Last edited: Oct 28, 2007
  4. Oct 28, 2007 #3
    I think that there is something wrong.

    The equations which you have applied is for a projectile motion on a horizontal plane.

    But here the rider is starting from a height of 15 m.

    And by the equations which you have applied it means that if the bike launches at a speed of 17.8 m/s he would be at the other end of the bank but at the height of 15m and hence its range would be greater than 40m.
  5. Oct 28, 2007 #4
    it's correct since it's also the answer at the back of the book. i took [tex]y_{0}[/tex] to be 0 which is at the point as he leaves the ramp and [tex]y=-15m[/tex] unless i lucked out.
  6. Oct 28, 2007 #5
    OK. It gets solved if you take

    [tex]y_0 = 0[/tex]


    [tex] y = -15[/tex]
  7. Oct 28, 2007 #6
    first i thought i would have to find the max height reached, and then go from there. i just find it strange that the descent isn't actually -15m, but i guess the rider barely goes above y=0 so it's negligible.

    physics is weird to me.
    Last edited: Oct 28, 2007
  8. Oct 28, 2007 #7


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    The point is that it doesn't matter how high above y=0 the rider travels; it is the total displacement that the equation depends upon. If he has a maximum height of 20m, say, he will have to come back down from 20m high, and so the net displacement of this motion will be zero.
  9. Oct 28, 2007 #8

    But the projectile is committed at a height of 15 m. Hence if we consider the initial height as 0 then we have to consider the final height as -15 m.
  10. Oct 29, 2007 #9


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    Perhaps I didn't explain it well enough, but I was merely trying to point out why any distance the rider goes vertically over the line y=0 did not add on to the total displacement.
  11. Feb 27, 2008 #10
    leaping the river

    I am working on this problem and have can't seem to find a solution. The answer needs to be in variable form and my answer doesn't seem accurate. Any help would be great.

    Leaping the River. A car of weight W_C car comes to a bridge during a storm and finds the bridge washed out. The driver of weight W_D must get to the other side, so he decides to try leaping it with his car. The side the car is on is h_1 above the river, while the opposite side is a mere h_2 above the river. The river is L wide.
  12. Feb 27, 2008 #11
    How fast should the car be traveling just as it leaves the cliff in order just to clear the river and land safely on the opposite side?

    What is the speed of the car just before it lands safely on the other side?
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