Will Ice Cream Land on Physics Professor's Head?

[gothloli]

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 happen? What is the relative speed of the ice cream with respect to the head at the time of impact?

Homework Equations

s = 1/2(u + v)t
and relative velocity
vectors

The Attempt at a Solution

I suppose that the vertical height as well as the horizontal distance is changing as the escalators go up and down. I'm confused as to whether I use relative velocity, since both escalators move at same velocity, but in opposite directions. I know that to find the time, you have to set their distance equations equal but they cancel, since velocity is the same. I attached file of the trig.

 

[frogjg2003]

There are two things I do not like about this problem:
1. It gives your professor's height, but not yours.
2. It doesn't say how the escalators are aligned with each other, except that they are adjacent (which I'm taking to mean parallel and connecting).

Even if you had a given height, you need to know if the horizontal velocities are parallel or antiparallel. If they are moving parallel, then you are directly overhead the professor, but if they are moving antiparallel, you are on opposite sides.

 

[Adonis1]

There are two things I do not like about this problem:
1. It gives your professor's height, but not yours.
2. It doesn't say how the escalators are aligned with each other, except that they are adjacent (which I'm taking to mean parallel and connecting).

Even if you had a given height, you need to know if the horizontal velocities are parallel or antiparallel. If they are moving parallel, then you are directly overhead the professor, but if they are moving antiparallel, you are on opposite sides.

Concerning #1, you are right. I guess one would have to either assume the ice cream drops from the top step, i.e. 10m, or make an assumption that it drops from a reasonable height at which it is being held, say 1m.

Concerning #2, the professor (not the one in the problem, but the one in the class for which this problem is assigned) mentioned in class that the escalator configuration should be interpreted as being like an X, and not looking like two parallel lines \\. In other words, you are directly overhead of the professor.

To Gothli: The ice cream's motion should be computed using ideal projectile motion no? I.e, it's x component is x=x0 + vx0(t) (where x0 is the ice cream's initial position horizontally, i.e. 0, and vx0(t) is the escalator's velocity 0.400 m/s), and its y component is: y=y0+vy0(t)-1/2gt^2 (where y0 is the height from which the ice cream drops, and vy0(t) is 0 since the y component is in free fall). So, in short, it seems to me in order to set the distance equations equal, we need to get the escalator velocity into both x and y components for the professor's motion don't we? I.e. for the Professor's motion, we can start the motion from the top of his head, but we need to know how far along each axis his head moves as a result of being directed by the escalator's motion. Well the vertical velocity should be 0.4cos(40) and the horizontal velocity should be 0.4sin(40).

 

[frogjg2003]

The ice cream's motion should be computed using ideal projectile motion no? I.e, it's x component is x=x0 + vx0(t) (where x0 is the ice cream's initial position horizontally, i.e. 0, and vx0(t) is the escalator's velocity 0.400 m/s), and its y component is: y=y0+vy0(t)-1/2gt^2 (where y0 is the height from which the ice cream drops, and vy0(t) is 0 since the y component is in free fall).

The ice cream is initially moving with the downward elevator, so wouldn't v_{0x}=v\cos(40^\circ) and v_{0y}=-v\sin(40^\circ)

 

[Adonis1]

The ice cream is initially moving with the downward elevator, so wouldn't v_{0x}=v\cos(40^\circ) and v_{0y}=-v\sin(40^\circ)

Yes that's right, thanks for the correction.

 

[gothloli]

Concerning #1, you are right. I guess one would have to either assume the ice cream drops from the top step, i.e. 10m, or make an assumption that it drops from a reasonable height at which it is being held, say 1m.

Concerning #2, the professor (not the one in the problem, but the one in the class for which this problem is assigned) mentioned in class that the escalator configuration should be interpreted as being like an X, and not looking like two parallel lines \\. In other words, you are directly overhead of the professor.

To Gothli: The ice cream's motion should be computed using ideal projectile motion no? I.e, it's x component is x=x0 + vx0(t) (where x0 is the ice cream's initial position horizontally, i.e. 0, and vx0(t) is the escalator's velocity 0.400 m/s), and its y component is: y=y0+vy0(t)-1/2gt^2 (where y0 is the height from which the ice cream drops, and vy0(t) is 0 since the y component is in free fall). So, in short, it seems to me in order to set the distance equations equal, we need to get the escalator velocity into both x and y components for the professor's motion don't we? I.e. for the Professor's motion, we can start the motion from the top of his head, but we need to know how far along each axis his head moves as a result of being directed by the escalator's motion. Well the vertical velocity should be 0.4cos(40) and the horizontal velocity should be 0.4sin(40).

oh my gosh, thanks so much. I'm such an idiot for not realizing question asking for when motion of ice cream falling meets the professor, I thought it was when the student comes across the professor, which wouldn't make sense. Yes it is ideal projectile motion. Thanks

The ice cream is initially moving with the downward elevator, so wouldn't v_{0x}=v\cos(40^\circ) and v_{0y}=-v\sin(40^\circ)

Thanks :)