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Projectile motion of a daredevil.

  1. Jan 17, 2007 #1
    G: The problem statement, all variables and given/known data.
    A daredevil jumps a conyon 12 m wide. To do so, he drives a car up a 15 degree incline.

    R:a. What minimum speed must he achieve to clear the canyon.
    b. If the daredevil jums at this minimum speed, what will his speed be when he reaches the other side?



    Attempted answer:
    I tried finding the value of the arch (hypeotenus) from the cliff to halfway over the cliff at the highest point. From there I found the cos of 15 times 6.19 to find 5.98. From there i tried sticking to one dimension, and using 5.98 as the velocity going down, and ovbiously -9.8 as acceleration. Then I used delta y=1/2at^2 + vt. I know that the answers to both are 15m/s, but i know my work is wrong. Help please! and if anyone has any tips to remember for projectile motion i would really appreciate it because i have a test on it soon.
     
  2. jcsd
  3. Jan 17, 2007 #2

    cristo

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    Staff Emeritus
    Science Advisor

    When attempting projectile motion questions, it is best to first split the motion into horizontal and vertical directions, and consider each one separately. Draw a diagram of the situation, and then draw up a table, one column for horizontal, one for vertical, and list the variables that you know.

    Then, look at the kinematic equations, and for each part, pick the suitable equation to give you what the question asks for.
     
  4. Jan 17, 2007 #3
    alright, tried breaking it up into x and y.

    v sin15=y
    v cos15= x

    x y
    delta x=1/2at^2+vt delta y= 1/2at^t + vt
    delta x=vt 0=1/2(-9.8)t^2=vt
    12=v*cos15*t 0=-4.9t^2+v sin15 t
    12.37/v=t 0=-4.9(12.37/v)^2 + v sin15 (12.37/v)
    749.8/v^2 = v sin15 (12.37/v)
    749.8/v^2 = 3.20v/v
    749.8 = 3.20v^2
    239.3125=v^2
    15.47

    aaaah! get it. lol. i was doing alg mistakes. thnx!
     
  5. Jan 17, 2007 #4
    aah! the work mixed up! oh well, i got it.
     
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