- #1

Nax

- 2

- 0

## Homework Statement

You are at the mall on the top step of a down escalator when you lean over laterally to see your 1.8 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 degrees with the horizontal, a vertical height of 10 m. 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

y

_{0i}+v

_{y0i}t-1/2gt^2

y

_{0p}+v

_{y0p}t

## The Attempt at a Solution

__I'm able to find time:__

10.0-(0.400)sin(40)(t)-1/2(9.81)t^2=1.80+(0.400)sin(40)(t)

-4.905t^2-0.25115t+10.0 = 0.25115t+1.8

-4.905t^2-0.51423t+8.2 = 0

(-4.905t^2-0.51423t+8.2)*-1 = 0 * -1

4.905t^2+0.51423t-8.2 = 0

t = -b +/1 sqrt ( b^2 - 4 ac) / 2a

t = ((-0.51423 +/- sqrt( (0.51423)^2 - 4(4.905) (-8.2) ) / (4.905)*2

t = (-0.51423 +/- 12.694425)/9.81

(-)

t = -13.20866/9.81

t = -1.35

(+)

t= 12.18019525/9.81

t= 1.24

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I'm not sure how to find the vertical height and relative speed with respect to the head at the time of impact. Insight would be greatly appreciated, thank you. :)