WHAT!? You’re telling me dance can be described through PHYSICS?
Yes! And here are some important concepts that you should be familiar with in order to understand why this is so:
Position: precisely where an object is located
x(t) = x0 + v0t + 0.5at
V0 = initial velocity
Velocity: the magnitude AND direction in which a object is moving (therefore it is a vector)
v = displacement/time (+direction!)
Displacement = change in position
Given constant acceleration: v = v0 + at
When no time given: vf^2=v0^2+2a(xf-x0)
Acceleration: how rapidly the velocity of an object is changing
a = change in velocity/time
Yes! And here are some important concepts that you should be familiar with in order to understand why this is so:
Position: precisely where an object is located
x(t) = x0 + v0t + 0.5at
V0 = initial velocity
Velocity: the magnitude AND direction in which a object is moving (therefore it is a vector)
v = displacement/time (+direction!)
Displacement = change in position
Given constant acceleration: v = v0 + at
When no time given: vf^2=v0^2+2a(xf-x0)
Acceleration: how rapidly the velocity of an object is changing
a = change in velocity/time
Fun Physics facts about Jumps
-Gravity is the only force acting on a dancer in mid-air.
-Gravity affects only VERTICAL (up and down) motion, NOT horizontal (left to right)!
Therefore… time in the air depends only on the VERTICAL component of velocity, not on the dancer's mass nor horizontal speed… We can see this in a graph:
Dancer’s Position vs. Time During a Jump:
-Gravity is the only force acting on a dancer in mid-air.
-Gravity affects only VERTICAL (up and down) motion, NOT horizontal (left to right)!
Therefore… time in the air depends only on the VERTICAL component of velocity, not on the dancer's mass nor horizontal speed… We can see this in a graph:
Dancer’s Position vs. Time During a Jump:
As we can see, gravity only influences the vertical component of motion, not the horizontal.
Do you understand the graph…?
If not, do not worry! Here are some helpful explanations:
-When v=0 in the vertical velocity graph, the dancer is at the highest point in the jump.
- In this jumping example, we will call “up” the positive direction.
-So the positive velocity (above the x-axis, before v=0) represents the dancer’s direction (she jumps up).
-The negative velocity (below the x-axis, after v=0) means that the dancer is moving in the opposite way relative to the chosen direction. Because we chose “up” as a positive direction, then when the dancer is going down from the jump, the velocity is negative.
*This negative velocity represents direction*
-The negative slope means that the dancer slows down as she jumps up, until she reaches the maximum height (v=0). After the dancer reaches maximum height and is going down, the velocity increases in the downward direction.
…Still don’t get it? Here’s an example!
Let’s say that right before landing, the dancer has a velocity of -3m/s. The negative indicates a direction (down) and the 3 indicates magnitude (3m/s).
If not, do not worry! Here are some helpful explanations:
-When v=0 in the vertical velocity graph, the dancer is at the highest point in the jump.
- In this jumping example, we will call “up” the positive direction.
-So the positive velocity (above the x-axis, before v=0) represents the dancer’s direction (she jumps up).
-The negative velocity (below the x-axis, after v=0) means that the dancer is moving in the opposite way relative to the chosen direction. Because we chose “up” as a positive direction, then when the dancer is going down from the jump, the velocity is negative.
*This negative velocity represents direction*
-The negative slope means that the dancer slows down as she jumps up, until she reaches the maximum height (v=0). After the dancer reaches maximum height and is going down, the velocity increases in the downward direction.
…Still don’t get it? Here’s an example!
Let’s say that right before landing, the dancer has a velocity of -3m/s. The negative indicates a direction (down) and the 3 indicates magnitude (3m/s).
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The jump can be seen easier in this graph representing a projectile path ------>
As you can see, the vertical component of velocity gradually decreases until it reaches the highest point (because of the downward acceleration due to gravity). As the dancer comes back down, the vertical velocity increases downward. Note that throughout this time, the horizontal component remains constant. |
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Think you got this all down?!
Let’s do a practice problem to see☺
A dancer wants to calculate her initial velocity. The time it took her to jump is 1 second. How would you go about to solve this…?
Find: initial velocity
Know: a=-9.81 m/s2 (a constant equal to the gravitational acceleration)
Let’s do a practice problem to see☺
A dancer wants to calculate her initial velocity. The time it took her to jump is 1 second. How would you go about to solve this…?
Find: initial velocity
Know: a=-9.81 m/s2 (a constant equal to the gravitational acceleration)
Plan:
* To find ending velocity , you want to find out the distance from the ground to the peak of your jump. As we saw in the graph, at the peak of the jump, v=0 (because you are no longer traveling upward, but it is before you begin traveling downward).
*Because it takes the same amount of time to go from the initial velocity to v=0 and from v=0 to the final velocity, we should cut time in half: t=0.5sec.
*Think of the maximum point as a line of symmetry. So at the same height, you have the same magnitude in velocity.
Calculate:
We plug the known information:
0= V0+ -9.81 * 0.5sec
Answer: V0 =4.91m/s = about 5m/s
* To find ending velocity , you want to find out the distance from the ground to the peak of your jump. As we saw in the graph, at the peak of the jump, v=0 (because you are no longer traveling upward, but it is before you begin traveling downward).
*Because it takes the same amount of time to go from the initial velocity to v=0 and from v=0 to the final velocity, we should cut time in half: t=0.5sec.
*Think of the maximum point as a line of symmetry. So at the same height, you have the same magnitude in velocity.
Calculate:
We plug the known information:
0= V0+ -9.81 * 0.5sec
Answer: V0 =4.91m/s = about 5m/s
Application
If you didn't know how to solve the previous problem, don't worry! Here is how to do it so you know next time:)
Find: Vertical Distance
Know:
*a=-9.81 m/s2 *t=0.5sec
*v=0 *V0 =4.91m/s
Plan: To solve, we can use the equation for vertical distance: y(t) = y0 + v0t + 0.5at2
Calculate:
Distance = 0 + 4.91(.500) + 0.5(-9.81)(.500) 2
Answer: 1.23meters = about 1m
WOW! That's a high jump!
Find: Vertical Distance
Know:
*a=-9.81 m/s2 *t=0.5sec
*v=0 *V0 =4.91m/s
Plan: To solve, we can use the equation for vertical distance: y(t) = y0 + v0t + 0.5at2
Calculate:
Distance = 0 + 4.91(.500) + 0.5(-9.81)(.500) 2
Answer: 1.23meters = about 1m
WOW! That's a high jump!
