Chapter 3 - 2D Kinematics Flashcards
When to use basic definitions of velocity, acceleration, etc?
If given componetns of these, don’t even worry about kinematic equations
Mistakes with projectile motion problems x3
- Remember that at max height, velocity equals 0
- Object travels its maximum distance, angle is 45 degrees
- Forgetting that vox = vx because there is no horizontal acceleration
Projectile Motion - Time of Flight Problems
- Use the equation y = voy + 1/2at^2
- Let y = 0 and factor out and divide by t, then isolate for the t that still remains
- You should be left with the equation
-2voy/a where voy is vosintheta - Input and solve baby
Projectile Motion - Range Problems
What do you need in order to sovle for range?
- Use the eqaution x = voxt + 1/2at^2
- Since acceleration is 0 in teh x direction (because of constant velocity), this reduces down to
x = vox(t)
Time of flight
Projectile Motion - Max Height Problems
- Use the equation v^2 = v0y^2 + ay
- V^2 = 0 at the maximum height and voy = vosintheta
- Isolate and solve for y
A daring swimmer dives off a cliff with a running horizontal leap, as shown
in Figure. What must her minimum
speed be just as she
leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below the top
of the cliff
Time is teh same for both the x and y component so calculate using y component becuase you have acceleration and you know inital velocity in the y direction is 0.
using y = voyt + 1/2at^2
Sub in this time value into the vox equation. This equals initial velocity because voy=0
Grasshopper leaps into air from edge of vertical cliff at an angle of 50 degrees from the horizontal. He leaps up 0.0674m and the width of his leap is 1.06m. find his intial speed and the height of the cliff
PART A
You can use the equation
vf^2 = vi^2 + 2ay and isolate and solve for vi, knowing that the vf = 0 at the peak
use the voy value to calcualte vo
you should get 1.5m
PART B
1. calculate vox and use this to calcualte the tiem with the range calculation
2. using this time, input nto
y = voyt + 1/2at^2
ensuring that you use a = -9.8
you shuld get -4.5692, which makes sense as the displacement is downwards. now this is the total height of the jump, but you have to subtract from the height he jumps initally to find that off teh cliff
you should get 4.60
How do you solve projectile motion problems when there is a change in displacement?
Remember that in two of the kinematic equations that have x in front, there is also x0 included in the equation.
Otherwise for example calculating max height, you can disregard it and add 1.5 at the end
A skier launches off a slope with an inital velocity of 8 directed at an angle of 30 degrees above the horizontal. top of slope is 1.5m off the ground.
a) how high above the ground is the highest point the skier reaches.
b) when the skier reaches the highest point, how far horizontally is this point from the top of the slope?
c) calculate the skier’s range
a) For the first part of this question, you want to set vf = 0 and solve for y using the eqatuion v^2 = voy^2 + 2ay
Don’t forget the squared, and you should get 0.816
b) To find the time to reach top you can use the equation
vf = voy + at
where vf = 0 and you should get t = 0.408 seconds
Then apply into the horizontal range equation
x = voxt and solve for x
c) Now you need to find the time of flight of the downwards part of motion
Use the equation y = y0 + v0t 1/2at^2
Inital velocity is at rest so you get
0 = 2.316 + 1/2(-9.8)t^2
and you get 0.686 seconds
Add that onto your original time of 0.408 seconds to get a cumulative of 1.094 seconds.
Then calcualte the range with this
x = voxt
you should get 7.57m
In a marathon race Chad is out in front, running due north at a speed of 4.00 m/s. John is 95 m behind him, running due north at a speed of 4.50 m/s. How long does it take for John to pass Chad
Relative speed - difference between their speeds.
Calculate the difference (0.5m/s) and then do time= d/s
A marble is thrown horizontally with a speed of 15 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 65° with the horizontal. From what height above the ground was the marble thrown?
Since there is no horizontal acceleration vox = vx. Thus can find v using the vox = vxcostheta
This value can then be used to find vy
Consequently, this value is subed into vy^2 = voy^2 + 2ay
where voy = 0 because object was thrown vertically
A diver springs upward from a diving board. At the instant she contacts the water, her speed is 8.90 m/s, and her body is extended at an angle of 75.0° with respect to the horizontal surface of the water. At this instant her vertical displacement is −3.00 m, where downward is the negative direction. Determine her initial velocity, both magnitude and direction.
You want to calculate the vfx and the vfy using these values.
You know that vfx = vox and you can find voy using the equation vy^2 = voy^2 + 2ay
Then use both of these components to find the velocity using pythagorean theorem and tan of the components y/x to find the angle
A player kicks a football at an angle of 40.0° from the
horizontal, with an initial speed of 12.0 m/s. A second player
standing at a distance of 30.0 m from the first (in the direction
of the kick) starts running at the instant the ball is kicked to
meet it. How fast must he run in order to catch the ball just
before it hits the ground?
You have to find the time of flight of the projectile using the vertical direction information that you have and the equation
y = vot + 1/2at^2
and you should get 5 = 1.57 sec
Then you sub into the equation for horizontal range to determine the horizontal range of teh projectile
x = voxt
you should get 12.11
THEN you subtract 30 - 12.11 to figure how much the boy has to fun in the 1.57sec, which will be your speed
