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Homework Help: Motion in 2D ball throw

  1. Mar 7, 2005 #1
    A ball is thrown toward a cliff of height h with a speed of 30 m/s and an angle of 60degrees above horizontal. It lands on the edge of the cliff 4.0 s later.What was the maximum height of the ball?
    I try using this equation:
    for y_i, i have 0
    for v_iy i have 25.5
    for (deltaT)i have 4s
    and a_y=9.8
    I am not getting the right answer, why not?
  2. jcsd
  3. Mar 7, 2005 #2
    [itex]v_f-v_0=g\Delta t[/itex] Solve for [itex]\Delta t[/itex].

    The t you are given in the original problem is not the same as it takes for the ball to reach its maximum height. Now use another kinematics equation involving height, initial velocity, acceleration, and time to find how high it goes. If you know the correct answer, post it here.
  4. Mar 7, 2005 #3
    didnt work...
  5. Mar 7, 2005 #4


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    What didn't work?

    The height of the cliff and the time it takes the ball to get to the cliff is irrelevant to the question.

    The only force on the ball is gravity- straight down. The only acceleration is 9.8 m/s2- straight down. The initial speed is 30 m/s at 60 degrees so the vertical component of initial velocity is 30sin(60)= 15√(3) m/s, At any time t, the vertical component of velocity is 15√(3)- 9.8t. The ball goes up as long as that is positive and down once it is negative- at its highest point, the vertical speed of the ball is 0: 15√(3)- 9.8t= 0 or t= 15√(3)/9.8. The height at anytime t is
    h= 15√(3)- 4.9t2. Put in t= 15√(3)/9.8 to find the highest point.
  6. Mar 7, 2005 #5
    "Put in t= 15√(3)/9.8 to find the highest point." t will equal 2.65s If i plus that back into h= 15√(3)- 4.9t2. i will get a negative answer. how is that the highest point?
  7. Mar 7, 2005 #6


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    One possibility here is that the initial height "y0" from which the ball is thrown is NOT equal to (0). In this case, the answer must be given in terms of Cliff Height "h":
    {Maximum Height of Ball} = y0 + Δy
    Since the ball experiences constant vertical acceleration "g" and has velocity=(0) at its max height, we can obtain "Δy" from:
    Δy = (vY0)2/(2g) = {30*sin(60o)}2/{2*(9.81)} = (34.4 m)
    so the first equation becomes:
    {Maximum Height of Ball} = y0 + (34.4 meters) :::: Eq #1
    From the problem, we know the ball reaches height "h" from an initial height "y0" after t=(4 sec), so we can write:
    h = y0 + (vY0)(4 sec) + (1/2)(g)(4 sec)2 =
    = y0 + 4*{30*sin(60o)} - (1/2)(9.81 m/sec2)(4 sec)2 =
    = y0 + (25.4 m)
    ::: ⇒ y0 = h - (25.4)
    Placing this result in Eq #1 above, we get:
    {Maximum Height of Ball} = {h - (25.4)} + 34.4
    {Maximum Height of Ball} = h + (10 meters)
    Thus, ball reaches Max Height of (10 meters) above Cliff.

    Last edited: Mar 7, 2005
  8. Mar 8, 2005 #7
    You're obviously just copying the answer from HallsofIvy. But, he accidentally forgot to put the time next to the velocity of the height equation, you took it, plugged it into the calculator, and got the wrong answer. You need to find the time it takes to reach a peak, and look for the correct equation.

    [tex]\Delta h=h_0 + v_0 t + \frac{1}{2}gt^2.[/tex] You know v, you know t, you know g.

    Xanthym has used an interesting method too, but I'm not sure it's easier. It looks to take more time than is neccesary, but it shows you that you can find more than one way to skin a cat. Just because something "doesn't work" doesn't mean you should give up and stop working on it.

    Good luck. :cool:
  9. Mar 8, 2005 #8
    max height H= (u sin theta)^2 /2g
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