multiple choice questions

free response questions.

Answers to 1-15

Multiple | choice | answers | 1 through | 15 | |

1) 2 |
2) 3 |
3) 1 |
4) 4 |
5) 1 |
6) 1 |

7) 1 |
8) 1 |
9) 3 |
10) 4 |
11) 2 |
12) 1 |

13) 4 |
14) 1 |
15) 3 |

Answers to free response questions:

1. Each ball falls vertically at the same rate, so each will take the same amount of time to reach the ground. This is because each starts with an initial vertical velocity of zero (viy = 0). The ball dropped from the moving car also has a horizontal velocity that remains constant. That ball will have a greater speed than the ball dropped from the car at rest (because it has a horizontal component to its speed) and it will travel a further distance (taking into account the horizontal distance as well as the vertical distance).

2) The horizontal acceleration is zero and the graph is thus a horizontal line at 0.

3) We know that at 45 degrees, the vertical and horizontal components are equal (because cos 45 = sin 45). If the angle is less than 45 degrees, the x-component will be greater than the y-component. Remember the demonstration with the three meter sticks -- as the angle decreases, the x-component increases while the y-component decreases.

4) The acceleration vector is straight down. The velocity vector will have the same angle at impact as the initial velocity (40 degrees), only at impact the arrow points down at 40 degrees.

5) **KINEMATIC EQUATIONS:**

avg. velocity = d/t

avg velocity = (vi + vf)/2

d = 1/2(vi + vf)t

vf = vi +at;

d = vi(t) + 1/2 at^2;

vf^2 = vi^2 + 2ad.

**5)** Initial velocity is 20 m/s; angle is 35 degrees. Find maximum height, time of flight and
range.

Follow the patented ** Byrne method** for projectile motion problems:

a) Diagram the problem. The projectile is represented by an arrow at a 35 degree angle above the horizontal. The magnitude of the arrow (the initial velocity, vi) is 20.0 m/s.

b) Create a coordinate system: Here, there is upward motion, so I will call up "positive" and down "negative", to the right will be "positive" and to the left "negative."

c) Create an x-/y- data table and fill in the known variables and solve for time on the y-side.

Use the equations Ax = Acos(theta), Ay = Asin(theta) to find the x- and y- components of the initial velocity:

y-components | x-components |

vi(y) = 20sin35 m/s = 11.5 m/s | vi(x) = 20cos35 m/s = 16.4 m/s |

a = -9.81 m/s2 | a = 0 |

vf = 0 (when the projectile is at its highest point) | |

t = ? | t= (we'll use double the time found on the y-side) |

d = ? | d= ? |

d) We now solve for t in the y- direction. We can use vf = vi + at.

Substitute with units: 0 = 11.5 m/s - 9.81 m/s^2(t)

t = 1.17 s

e)Now find d using any equation with d in it. I will use: d = vit + 1/2at^2

Substitute with units: d = (11.5 m/s)(1.17 s) - .5(9.81 m/s^2)(1.17 s)^2 (I took the minus sign and placed it before the .5)

d = 6.74 m

f) We now need the time of flight. We simply double the time we found above because the time
it takes to go up to the highest elevation is half the time of flight. So time of flight equals 2.34 s.

g) To find the range, we will carry the time over to the x- column. We will find the x- displacement using d = (avg. v)t,
the only equation we can use on the x-side. Remember, because the acceleration on the x-side is zero, vi on the x-side
is the same as vf on the x-side. So vix is average velocity on the x-side.

Our data table now looks like this:

y-components | x-components |

vi(y) = 20sin35 m/s = 11.5 m/s | vi(x) = 20cos35 m/s = 16.4 m/s |

a = -9.81 m/s2 | a = 0 |

vf = 0 (when the projectile is at its highest point) | |

t = 1.17 s | t= 2.34 s |

d = 6.74 | d= ? |

g) We will solve for d in the x direction, using d = average velocity times time (vt).

Substitute with units: d = (16.4 m/s)(2.34 s) = 38.38 m.

Create your own problems and check your answers using the
Projectile Motion Applet.

Practice your motion graphs using Motion
Graphs for Constant Acceleration.