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Tennis ball crossbow calculations

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Ok, for my Mechanical engineering class we have been assigned to make a 2ft ^3 5kg device that will launch a tennis ball 20-40 ft. My group decided to use a design more or less like a crossbow, so I was hoping someone would look over my calculations and make sure they're right. I'm doing a rough calculation to make sure the bungee cords we use are strong enough to reach the max distance.

    Here's what I've got.
    Assume a tennis ball ways about 57g, will be accelerated over 1.75 ft at 45 degrees from horizontal, on a vertical flat surface

    2. Relevant equations
    F=ma
    Vf= [itex]\sqrt{2a \Delta X}[/itex]
    a2+b2=c2
    -Vx and Vy make a right triangle with V

    Vy+ g*t= -Vy
    since it's height is the same at launch and impact it's velocity should be the opposite of it's initial.
    2Vy = -g*t

    3. The attempt at a solution
    sorry, the itex started breaking on me, and I couldn't figure out where...
    Vf=sqrt(2(F/56.7g) * 1.75ft)= sqrt(Vx^2 + Vy^2)
    Vx=Vy
    t = 40ft/Vx

    2(F/56.7g)*1.75ft=2Vy^2=-gt
    2(F/56.7g)*1.75ft=2Vy^2=-g*40ft/Vy
    Vy^3=-(-9.81m/s^2)*40ft /2
    Vy^3=(9.81m/s^2)*20ft
    Vy=(3.911m/s)
    2(F/56.7g)*1.75ft=2(3.911m/s)^2
    F=1.626 newtons, and this would need to be the average force over the time of acceleration, so the initial force would be 2x because at the end position it would be 0 force so (2x +0)/2= x, thus 3.2 N ?

    Thanks in advance.
     
    Last edited: Nov 22, 2011
  2. jcsd
  3. Nov 22, 2011 #2
    Seems like that is not enough...


    using these equations I would solve each for v,
    2Vy = -g*t
    t = 40ft/Vx

    then plug them into Vx=Vy(also I would convert everything to the same unit system, i.e. N, m, kg) you could then get t and plug it back in for v
     
  4. Nov 23, 2011 #3
    You have determined the Vx and Vy velocity components. You should do some checking.

    Vx = 3.911 m/s = 12.83 ft/s

    For the time of flight (up and down from launch point) you have the following quadratic. Solve it for t. The -1.237 comes from the fact the ball lands that far below where it was released.

    -1.237= 12.83*t-16.1*t*t

    With the t value, plug into Sx=Vx*t to see if the ball has traveled the 40 feet. If you don't get 40 feet, you have an error.
     
  5. Nov 29, 2011 #4
    ok, using the quadratic you set up I set it as -1.237= Vy*t -16.1 t^2, and t=40ft/Vy, and I got Vy=25ft/s, Vf=sqrt(2*(25ft/s)^2)=35.35ft/s
    V^2=2a(1.75)
    V^2 /2.5 = a = 499.7ft/s^2
    F=ma = 499.7ft/s^2 * 57g = 8.6816N = 1.95lbs
    Am I correct in assuming I should double this to find the initial force because as the spring returns to it's natural state the force it is applying will go to zero, right? So the initial force should be about 3.9 lbs because F(avg) = (Fi+Ff)/2 = (Fi+0)/2

    Thanks again for your help, I didn't think of looking at it that way, and also didn't consider the height of the ball when it's released.
     
  6. Nov 29, 2011 #5
    When I solve the above quadratic, I get .88 seconds for the positive root. The tennis ball will have only travelled 12.83 * .88 = 11.3 feet. To go 40 feet, you'll need a velocity greater than 35 ft/sec.

    You will have to construct a device, like a crossbow, that can store enough potential energy when compressed to propel a 57 gram ball at a speed greater than 35 ft/sec in a distance of 1.75 feet. But your are not done yet. The ball is being accelerated up the ramp so you must account for the potential energy change in the ball as well.

    The energy required will be .5mv^2+mgdh where dh is the rise of the ramp.
     
    Last edited: Nov 29, 2011
  7. Nov 29, 2011 #6
    ok, so
    1/2 * 57grams * (35ft/s)^2 + 57grams*sqrt(1.75ft^2)/2*9.81m/s/s = 3.4562 joules
    so to release this amount of energy as the ball travels the 1.75ft barrel length, the avg. force applied will need to be 6.292 newtons, and then the initial force would be 12.58 newtons or 2.83lbs? (neglecting friction, and the Force of gravity on the spring, and assuming the spring is at it's natural length when the ball reaches the end of the crossbow.)
     
  8. Nov 30, 2011 #7
    That's numerically close. I compute 6.46 N. I do not agree that the height is 1.75 feet. Use trigonometry to determine the correct height for your change in potential energy. While in this problem it is a small portion of the total energy imparted to the ball, it nevertheless should be corrected. You are mixing units and I suspect this leads to your inaccuracy.
     
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