Ballistics Demonstrator

The science of ballistics involves the motion of projectiles. Exterior ballistics concerns the projectile after it leaves the device that it is shot from. The term "ballistics" comes from the name of one kind of catapult, the ballista.

Achievement

  1. Identify the Problem
  2. Generate Concepts
  3. Develop a Solution
  4. Construct and Test a Prototype
  5. Evaluate the Solution
  6. Present the Soltution

Constraints

Standards

Engineer's Notebook—10 points

Evaluated using the usual rubric. Make sure you:

The Device—10 points

The device will be judged based on the quality of construction and the design elements that let it shoot a marshmallow a consistent and repeatable distance.

Performance During Calibration—20 points.

5 points each for:

Performance at the Target—20 points.

Your score is 20 points minus your total distance away from the target in feet.

Completition of the Problems—20 points

Make sure and:

Ballistics Formulae (see derivation below)

Range X = Vi2sin2θ/(-g)

Altitude Ymax = Vi2sin2θ/(-2g)

Time of Flight tof = 2Visinθ/(-g)

The Problems

  1. (3 points) A skydiver jumps from 10,000 ft. How fast is she going after:
    a. 1 second?
    b. 2 seconds?
    c. 8 seconds?
    d. What is her altitude after 8 seconds?
     
  2. (3 Points) A marshmallow is dropped from the roof of the school, 65 feet from the ground. How fast is it going and what is its altitude after:
    a. 1 second?
    b. 2 seconds?
    c. 3 seconds?
    d. What is her altitude after 8 seconds?
     
  3. (4 Points) A baseball is thrown straight up at 64.4 ft/sec.
    a. How long does it take to get to its highest point?
    b. How high does it go?
    c. How long does it take to get back to your glove?
    d. How fast is it going when you catch it?
     
  4. (3 Points) A projectile is thrown at 100 ft/sec at an elevation angle of 45°.
    a. How far does it go (range)?
    b. What is the maximum altitude?
    c. What is the time of flight?
     
  5. (3 Points) A projectile is thrown at 100 ft/sec at an elevation angle of 30°.
    a. How far does it go (range)?
    b. What is the maximum altitude?
    c. What is the time of flight?
     
  6. (4 Points) A projectile is thrown at 100 ft/sec at an elevation angle of 60°.
    a. How far does it go (range)?
    b. How does this range compare to the range at 30?
    c. What is the maximum altitude?
    d. What is the time of flight?
    Bonus: at which angle did the projectile go farthest? Why don't hit baseballs fly the farthest at this angle?
     

    External Ballistics Formulae

    General Equations Under Constant Acceleration

    Acceleration (a) = constant = g = -32.2 ft/sec2 or -9.8 m/sec2

    Velocity is then equal to the intitial velocity (Vi or sometimes V0) plus the acceleration times time.

    Velocity (V) = Vi + gt

    Position is the initial position (d0), plus the initial velocity times time, plus one half acceleration times time squared.

    Position (d) = d0 + Vit + ½gt2

    External Ballistics Input Variables

    The range, altitude, etc. of a projectile depend only on the initial velocity (Vi or sometimes written V0) and the elevation angle (Greek Theta, θ)

    The initial velocity vector can be broken into its X and Y components where:

    Velocity in the Y (vertical) direction = Visinθ

    Velocity in the X (horizontal) direction = Vicosθ

    Initial Conditions

    In the Y (vertical) direction the equations become:

    Velocity (VY) = Visinθ + gt

    Altitude (Y) = Y0 + Visinθ t + ½gt2

    In the X (horizontal) direction there is no acceleration. Assume the projectile starts at the origin. It will move at a constant speed in X and then:

    Velocity (VX) = Vicosθ

    Position (X) = Vicosθ t

    Dropping a Ball (or other projectile)

    The initial velocity is 0 and the ball accelerates downward, gaining 32.2 feet per second of downward velocity each second.

    Velocity (VY) = gt

    The altitude will be the original altitude (Y0) plus one half g times time squared. But since g in negative, the altiude will go down.

    Altitude (Y) = Y0 + ½gt2

    Note that g is a negative number and that the time is squared in the altitude formula.

    Throwing a Ball Straight Up

    In this case, the initial altitude (Y0) is 0. The ball will go up, slowing down until it stops at the top. The ball will then accelerated downward and return to your hand with same speed that it left, but in the opposite direction. The initial velocity is Visinθ, but since the sine of 90° is 1, the initial velocity is just Vi

    How long does it take to get to the top? At the highest altitude V = 0

    V = 0 = Vi + gt

    -gt = Vi

    ttop = Vi/(-g)

    What is the maximum altitude (Ymax)? Put the time to the top in the altitude equation.

    Altitude (Y) = Y0 + Vi t + ½gt2

    Ymax = Y0 + Vi t + ½gt2

    Set Y0 to 0 and replace t with Vi/(-g)

    Ymax = Vi Vi/(-g) + ½g[Vi/(-g)]2

    Ymax = Vi2/(-g) + ½[g/(-g)2]Vi2

    Ymax = -Vi2/g + ½Vi2/g

    Ymax = ½Vi2/(-g) or Vi2/(-2g)

    In the general case, ball not thrown straight up:

    ttop = Visinθ/(-g)

    Ymax = Vi2sin2θ/(-2g)

    Time of Flight

    When the ball gets back to your hand the altitude is once again 0.

    Altitude (Y) = Y0 + Vi t + ½gt2

    0 = 0 + Vi t + ½gt2

    Factor out t on the right

    0 = t [Vi + ½gt]

    This equation has two solutions, one then t = 0 and then when:

    0 = Vi + ½gt

    -½gt = Vi

    And the time of flight tof = 2Vi/(-g), which is twice the time to the top

    In the general case, ball not thrown straight up: tof = 2Visinθ/(-g)

    Launching a Projectile at an Angle. What is the range?

    There is no acceleration in X. If you set X0 to 0, the equation becomes:

    X = Vicosθ t

    Substituting the time of flight for t

    X = Vicosθ2Visinθ/(-g)

    Rearranging

    X = Vi22cosθsinθ/(-g)

    But 2cosθsinθ is equal to sin2θ

    Range (X) = Vi2sin2θ/(-g)

    In Summary

    Range X = Vi2sin2θ/(-g)

    Altitude Ymax = Vi2sin2θ/(-2g)

    Time of Flight tof = 2Visinθ/(-g)

    Here is a Table of Common Sines and Cosines

    Anglesincossin2cos2sin2θ
    30°0.50.870.250.750.87
    45°0.7070.7070.50.51
    60°0.870.50.750.250.87
    90°10100

     


    Links

    Fat Brain Catapult Kit
    The Grommet
    Don't try this
    And really don't try this
    Or even this
    And definitely not this



    Online Physics Games

     

    Last updated on 22 June 2018 by P. A. Wiedorn

     

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