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

  • Thread starter Thread starter maxtheminawes
  • Start date Start date
  • Tags Tags
    Sky
AI Thread Summary
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.
maxtheminawes
Messages
20
Reaction score
0

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
 
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.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top