I believe I can Fly~I believe I can touch the sky

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Homework Help Overview

The problem involves projectile motion, where a character attempts to jump off a cliff at an angle while running. The discussion focuses on calculating various aspects of his motion, including maximum height, velocity at that height, time in the air, final velocity, and horizontal displacement.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the separation of horizontal and vertical components of motion, questioning the nature of horizontal acceleration and its implications for the problem.

Discussion Status

The discussion is ongoing, with some participants clarifying concepts related to horizontal motion and acceleration. There is an exploration of the assumptions regarding forces acting on the projectile.

Contextual Notes

Participants are considering the effects of gravity and the lack of horizontal forces in the context of projectile motion, while also addressing the implications of these assumptions on the calculations presented.

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Homework Statement


Augustine continues to run around. He gets to the park and has the overwhelming urge to jump off a cliff and try to fly. He does so at an angle of 50 degrees to the horizontal while running at 14.5m/s. With his arms flapping, he rises in the air.
a) what is the highest point from the top of the cliff that he reaches before the laws of physics catch up with him and start bringing him back down to Earth?
b)What is his velocity at that point?
c)If the cliff is 10m high, how long is he in the air before he comes crashing down to the nice, soft, sandy beach below the cliff?
d)what is his final velocity?
e)what is his final horizontal displacement from the cliff?




Homework Equations


v=vo+at
y=yo+volt+1/2at^2
x=xo+volt+1/2at^2
v=final velocity
vo=initial velocity
yo=initial y displacement
xo=inital x displacement


The Attempt at a Solution


a) Vx=14.5cos50=9.32
Voy=14.5sin50= 11.11

v=vo+at
0=(9.32)+(-9.8)t
-9.32=-9.8t
-.95=t
.95sec=t

y=yo+volt+1/2at^2
y=11.11(.95)+1/2(-9.8)
y=1.23m


b)1.8m/s


c).95secs


d)v=vo+at
v=14.5+(-9.8)(.95)
v=5.59m/s


e)x=xo+volt+1/2at^2
x=14.5(.95)+1/2(-9.8)(.95^2)
x=23.85m
 
Physics news on Phys.org
Distinguish between horizontal and vertical components of acceleration, velocity and displacement.
What is horizontal component of acceleration? ehild
 
i really have no clue. is it constant. i know that acc downwards is -9.8m/s^2
 
Yes, and (ignoring friction) that constant is 0! There is no horizontal force and so no horizontal component of acceleration.
 

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