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

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Augustine attempts to fly by jumping off a 10-meter high cliff at a 50-degree angle while running at 14.5 m/s. The calculations reveal that he reaches a maximum height of 1.23 meters above the cliff before descending. His velocity at the peak is 1.8 m/s, and he remains in the air for approximately 0.95 seconds. Upon landing, his final velocity is 5.59 m/s, and he achieves a horizontal displacement of 23.85 meters from the cliff. The discussion emphasizes that the horizontal component of acceleration is zero, as there are no horizontal forces acting on him.
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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+vot+1/2at^2
x=xo+vot+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+vot+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+vot+1/2at^2
x=14.5(.95)+1/2(-9.8)(.95^2)
x=23.85m
 
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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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