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Homework Help: Throwing a ball and calculating the speed

  1. Sep 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A person with a 3 kg ball at rest in his hand, throws it with constant force at an angle of 45 degrees. The ball moves in a straight line until it leaves his hand at 2.1 meters above the ground. His arm has the length of 0.55 meters. Between the ball leaving his hand and until impact the total time elapsed is 2.15 seconds and the horizontal distance is 16 meters.

    What is the speed of the ball the moment it leaves his hand?

    2. Relevant equations
    Constant acceleration on a straight line:




    Projectile motion:
    [itex]x=(v_0 \cos(\alpha_0))t[/itex]

    [itex]y=(v_0 \sin(\alpha_0))t-\frac{1}{2}gt^2[/itex]

    [itex]v_x=v_0 \cos(\alpha_0)[/itex]

    [itex]v_y=v_0 \sin(\alpha_0)-gt[/itex]

    3. The attempt at a solution
    The text tells me that from the ball is at rest and until it leaves his hand, the movement is a straight line. Therefore I'm thinking I should use one of the equation for straight line.
    Also, if the force is constant it means the acceleration is constant as well (can somebody please confirm this to me?)

    So I want to find [itex]v_x[/itex]. I rearrange my coordinate system so that the straight line movement his arm makes is parallel to the x-axis.

    Thus the distance that the ball travels in this straight line is the length of the guy's arm, which is 0.55 meters so:
    [itex]x-x_0=0.55 m[/itex]

    I also know that since the ball is initially at rest:

    But I don't know the acceleration during the straight line motion nor do I know the time it takes for the ball from rest until it leaves his hand. Therefore I have no idea which equation to use.
    Maybe I should be looking at the projectile motion equations instead?
  2. jcsd
  3. Sep 20, 2013 #2


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    Staff: Mentor

    It seems that you are given a lot more information here than is required to determine the answer. All you are asked to find is the speed of the ball at the start of its free flight: given the time of free flight, distance travelled, etc., find its initial speed. What happens prior to the moment it leaves his hand seems irrelevant.
  4. Sep 20, 2013 #3
    There are more questions to the text but I just included everything to be on the safe side. :)

    So are you saying this is the answer:
    [itex]16m=(v_0 \cos(45deg))*2.15s[/itex]

    But the thing is that the ball is in projectile motion 2.1 meters above the ground and then lands 16 meters on the ground, so shouldn't I account for those 2.1 meters in some way?
  5. Sep 20, 2013 #4


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    Looks right.
    No. If an object has a horizontal component of velocity v for time t then the horizontal distance moved is vt. What's happening in the vertical at the same time is irrelevant. You next need to use the time calculated and the 2.1m to compute the vertical component of the launch velocity.
  6. Sep 20, 2013 #5
    I just find it strange that I'm not using the initial height of the ball and the length of the person's arm for anything.

    I calculated the highest point that the ball reaches in motion.
    The time:

    [itex]0=10.5m/s \sin(45deg)-9.8m/s^2*t[/itex]

    The height:
    [itex]y=(10.5m/s \sin(45deg))*0.76s-\frac{1}{2}*9.8m/s^2*(0.76s)^2[/itex]

    I'm also supposed to calculate the speed when it hits the ground. I have been given in the text that it hits the ground at 2.15 s

    I calculate the x and y components for the velocity:
    [itex]v_x=10.5m/s \cos(45deg)=7.4m/s[/itex]
    [itex]v_y=10.5m/s \sin(45deg)-9.8m/s^2*2.15s=-13.7m/s[/itex]

    But again I didn't use the 2.1 m height or the 0.55 m length of his arm.
    Have I done something wrong?
  7. Sep 20, 2013 #6


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    The length of arm, and all info pertaining to the time before release, must be for later questions. But you are right that there is soemthing wrong with the question. You have too much data on what happens after release, and it conflicts. From the given flight time and horizontal distance the launch speed must be about 7.5m/s horizontally. Since it's at 45 deg that's also the vertical speed. The vertical displacement in that time is the horizontal displacement less gt2/2, or about 16-20 =-4 m. That doesn't fit with the -2.1 m given. maybe this is in an environment where g has a different value.
  8. Sep 20, 2013 #7
    I have a last question that asks me to draw a free body diagram during the throw. Perhaps that's where the length of his arm and the height when the ball leaves his hand comes into play, right?
  9. Sep 20, 2013 #8


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    The length of the arm, yes, but not the launch height. Because the question provides conflicting data for the trajectory after launch you can get different answers according to which subset of data you use.
  10. Sep 20, 2013 #9
    But my methods for finding the initial speed, the highest point in the trajectory and the speed when it hits the ground, were all correct right? regardless of the answers?
  11. Sep 20, 2013 #10


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