Will the Ice Cream Land on Your Professor's Head? Exploring Relative Motion

In summary, when an ice cream falls off of a cone and onto a professor's head, the ice cream's speed and the professor's head's speed are both constant. The ice cream will land on the professor's head 15.946 seconds after falling, at a vertical height of 10.0 meters.
  • #1
NATURE.M
301
0

Homework Statement



You are at the mall on the top step of a down escalator when you lean over laterally to see your 1.80 m tall physics professor on the bottom step of the adjacent up escalator. Unfortunately, the ice cream you hold in your hand falls
out of its cone as you lean. The two escalators have identical angles of 40.0° with the horizontal, a vertical height of 10.0 m, and move at the same speed of 0.400 m/s. Will the ice cream land on your professor’s head? Explain. If it does land on his head, at what time and at what vertical height does that hap- pen? What is the relative speed of the ice cream with respect to the head at the time of impact?

Homework Equations



Idea projectile motion:
Y=Y[itex]_{}(o)[/itex] + V[itex]_{}(yo)[/itex]t-1/2gt^2
X=X[itex]_{}(o)[/itex]+V[itex]_{}(x)[/itex]t


The Attempt at a Solution



So for ice cream:
V[itex]_{}(xo)[/itex]=Vcosθ
V[itex]_{}(yo)[/itex]=-Vsinθ
Then,
Y[itex]_{}(ice cream)[/itex]= 10-Vsinθt-1/2gt^2

And for the professor:
Y[itex]_{}(prof)[/itex]=1.8+Vsinθ-1/2gt^2

Then equating Y[itex]_{}(ice cream)[/itex]=Y[itex]_{}(prof)[/itex], and solving for t,
I obtain t=15.946s. This value for time sees much to large. Any suggestions for what I may be doing wrong.
 
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  • #2
The prof is not falling downwards; He's riding the escalator.
 
  • #3
gneill said:
The prof is not falling downwards; He's riding the escalator.

Ok so then for the prof, I would remove the gravitational acceleration term, and assume he moves at constant speed while riding the escalator.

And just to clarify, when computing the relative speed of the ice cream, I can use
v=√[v(o)^2-2g(y-y(o)] right?
 
  • #4
NATURE.M said:
Ok so then for the prof, I would remove the gravitational acceleration term, and assume he moves at constant speed while riding the escalator.

And just to clarify, when computing the relative speed of the ice cream, I can use
v=√[v(o)^2-2g(y-y(o)] right?

Hmm. Is that v meant to be the relative speed? If so, I don't think so. The ice cream starts with a negative velocity and picks up speed as it falls. The Prof's head is moving upwards at a constant speed (equal in magnitude to the initial speed of the ice cream, in fact). So I'd expect the relative speed to be a sum of magnitudes.
 
  • #5
gneill said:
Hmm. Is that v meant to be the relative speed? If so, I don't think so. The ice cream starts with a negative velocity and picks up speed as it falls. The Prof's head is moving upwards at a constant speed (equal in magnitude to the initial speed of the ice cream, in fact). So I'd expect the relative speed to be a sum of magnitudes.

So, then I'd have v(rel.)=v(prof) + v(ice.c), where in this case v(ice.c)=√(v(x)^2+v(y)^2), and similar procedure for v(ice.c).
Though, I'm unsure whether we should maybe take only the y-compents of v(prof) and v(ice.c) , and then add them to obtain v(rel.)?
 
  • #6
Since both the prof and the ice cream share the same x-velocity, the x-velocities won't contribute to the relative speed. Just work with the y-speeds.
 
  • #7
gneill said:
Since both the prof and the ice cream share the same x-velocity, the x-velocities won't contribute to the relative speed. Just work with the y-speeds.

Alright thanks a lot gneill!
 

Related to Will the Ice Cream Land on Your Professor's Head? Exploring Relative Motion

1. What is relative motion?

Relative motion refers to the motion of an object in relation to another object or observer. It takes into account the movement of both objects and how they interact with each other.

2. How is relative motion different from absolute motion?

Absolute motion refers to the motion of an object in relation to a fixed reference point, such as the Earth. Relative motion, on the other hand, takes into account the movement of both objects and their interactions with each other.

3. What is the relative velocity?

The relative velocity is the velocity of an object in relation to another object or observer. It takes into account the motion of both objects and their interactions with each other.

4. How do you calculate relative motion?

To calculate relative motion, you need to determine the velocities of both objects and their direction of motion. Then, you can use vector addition to find the relative velocity of the objects.

5. What are some real-life examples of relative motion?

Some real-life examples of relative motion include a person walking on a moving train, two cars driving on the same road, and a bird flying while a plane is in motion.

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