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Heights of two balls thrown at the same time

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown at angle [itex]\Theta[/itex] and another ball is thrown at an angle (90-[itex]\Theta[/itex]) with the horizontal from the same point with velocity 39.2 m/s. The second ball reaches 50 m higher than the first. Find the heights.


    2. Relevant equations

    v = u -gt where v is the final velocity and u the initial velocity ( u = 39.2 m/s and v = 0)

    S = ut - 1/2 gt2

    3. The attempt at a solution

    Since the angles were not given, I did not find the vertical and horizontal component of velocity. I tried to find the time taken to reach the maximum height using v = u - gt and got the value as 4. I then substituted this value in the second equation and got the answer as 78.4 m. Since it says that one ball reached 50 m higher, I should either add or subtract 50 from this. Can I assume that this height is of the second ball ? Why not the first ? Even assuming that this is for the second ball, I get the answer as 28.4 and 78.4 m whereas the answer is 14.2 and 64.2.

    I then attempted to solve the problem using the second equation by assuming the heights as S and S+50 with u remaining the same. However t would vary and the equation was not leading anywhere.

    Where did I go wrong?
     
  2. jcsd
  3. Sep 10, 2011 #2

    kuruman

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    It is to be assumed that both balls have the same initial speed, 39.2 m/s otherwise you cannot solve the problem. If the angles are not given, just use θ, but you cannot assume that the y-component of the velocity of any ball is 39.2 m/s. This would mean it is shot straight up which is just not true. Write the kinematic equation 2a(Δy)=vy2-v0y2 for each ball and see what you get.
     
  4. Sep 10, 2011 #3

    vela

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    You're trying to solve for θ, so you want to eventually end up with an equation that has θ in it, which you can solve. In other words, as kuruman said, "If the angles are not given, just use θ," so you'll end up with an equation with θ in it.

    Your instructor has probably recommended you try to solve problems algebraically first and plug numbers in at the end. One of the reasons this is a good idea is to avoid the mistake you've made. Just because you can't calculate a number for vy right at the beginning doesn't mean anything is wrong. It just means you need to leave it in terms of θ, which is exactly what you want because you want to end up with an equation with θ in it.
     
    Last edited: Sep 10, 2011
  5. Sep 10, 2011 #4
    Since time is not given, I have to use the formula v2 = u2 - 2aS.

    For the first ball, v i.e. final velocity is 0 and u initial velocity is 39.2 Sin[itex]\Theta[/itex] m/s and a is 9.8 m/s2 and S is the distance.

    1536.64 Sin2[itex]\Theta[/itex] = 19.6 S

    For the second ball,

    1536.64 Sin2(90-[itex]\Theta[/itex]) = 19.6 (S+50).

    So I get two equations,

    19.6S = 1536.64 Sin2[itex]\Theta[/itex] ------- A

    19.6S + 980 = 1536.64 Sin2(90 - [itex]\Theta[/itex]) -------- B

    A + B

    39.2S + 980 = 1536.64 Sin 90

    39.2S = 1536.64 - 980 = 556.64

    S = 14.2 m and S+50 is 64.2 m.

    Is my working correct ?
     
  6. Sep 11, 2011 #5

    kuruman

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    It looks correct, but why do you say that sin2θ+sin2(90o-θ) = sin90o? Is that always true? I know the answer, but I want to make sure that you do too.
     
  7. Sep 11, 2011 #6

    That is why I was unsure whether my working was correct. Can you give me a hint ?

    Is there any other way to arrive at the answer ?
     
  8. Sep 11, 2011 #7

    kuruman

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    Use the trigonometric identity sin(90o-θ)=cosθ and see what you get.
     
  9. Sep 11, 2011 #8
    Sin2[itex]\Theta[/itex] + cos2[itex]\Theta[/itex] = 1

    I am yet to learn the trig identities fully, thanks for pointing it out.
     
  10. Sep 11, 2011 #9

    kuruman

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    An important trig result that you will have to use over and over again is "If two angles add up to give 90o, the cosine of one is the same as the sine of the other." Try to remember that. :wink:
     
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