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anonymous
 one year ago
A physics class launches their physics teacher off a building by first throwing her onto a trampoline only to be bounced off the roof. The teacher leaves the trampoline going 12m/s into the air, and onto a collection of cardboard boxes 45m below.
a) How high did the teacher bounce?
b) How long was the teacher in the air?
c) What was the teachers velocity just before she hit the boxes below?
anonymous
 one year ago
A physics class launches their physics teacher off a building by first throwing her onto a trampoline only to be bounced off the roof. The teacher leaves the trampoline going 12m/s into the air, and onto a collection of cardboard boxes 45m below. a) How high did the teacher bounce? b) How long was the teacher in the air? c) What was the teachers velocity just before she hit the boxes below?

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aaronq
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443406379306:dw because they dont give a horizontal velocity, we assume the bounce is straight up (only in the vertical direction). We have the acceleration due to gravity opposing the movement of the person, who slows down, comes to an instantaneous stop and begins moving down. We wanna know the highest point, which can be done in several ways, i'm assuming they want you to use kinematic equations of motion. We can use: \(\sf v_f^2=v^2_i+2a\Delta y\) Solving for the distance, \(\Delta y\), we get: \(\sf \Delta y=\dfrac{v_f^2v^2_i}{2a}=\dfrac{0^2(12~m/s)^2}{9.8~m/s^2*2}=7.34693~m\approx 7.3~m\) Try using the rest of the equations of motion to get b) and c).

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1Qu are we modelling the teacher as a point mass? [that was a physics joke, and therefore very unfunny]
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