anonymous
  • anonymous
Physics stuff!! A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as –32 feet per second squared and approximate your answer to 3 decimal places.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@ganeshie8 @nincompoop @pooja195 @Preetha any ideas?
Empty
  • Empty
I believe there are two ways to do this 1) Calculate the trajectory 2) Calculate the energy change
ganeshie8
  • ganeshie8
presuming the ball lands on the ground and not on the cliff... even though it was thrown vertically upwards

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anonymous
  • anonymous
this is a calc I problem by the way
ganeshie8
  • ganeshie8
maybe the angle shouldn't matter here because the ball gains its speed by the time it reaches cliff. You may start with the given acceleration due to gravity : \[v(t)-v(0)=\int\limits_0^t a(t)\,dt \]
anonymous
  • anonymous
okay s v(0) = 40 but how do we find v(t) ?
Empty
  • Empty
Continue further, \[v(t) = \frac{d x(t)}{dt}\]
anonymous
  • anonymous
wait slow down , how did you find s(t) ?
ganeshie8
  • ganeshie8
\[\begin{align}v(t)-v(0)&=\int\limits_0^t a(t)\,dt \\~\\ &=\int\limits_0^t -32\,dt \end{align}\] you're given that \(a(t)=-32\), see if you can integrate above now..
anonymous
  • anonymous
It is position. 500ft cliff above the ground=+500 Thrown upward with speed of 40 feet per second-it will eventually fall down to the ground which is negative=40t acceleration is negative because velocity is negative -32 feet per second squared t is seconds so second squared=-32t^2
anonymous
  • anonymous
v(t) - 40 = (-32t^2)/2 ?
anonymous
  • anonymous
Position s=0 means it is a at ground. You find t to know the time it took to reach the ground. You then used the velocity function by taking derivative of position. Plug that time it reached the ground to the velocity function.
ganeshie8
  • ganeshie8
nope, whats antiderivative of constant ?
anonymous
  • anonymous
well you add a variable in this case t
anonymous
  • anonymous
oh woops i see what i did wrong
anonymous
  • anonymous
v(t) = -32t + 40: also @Shalante if we take your s(t) and we derive it until we get a(t) we get -64 and i think it should be -32
anonymous
  • anonymous
no it is -64
anonymous
  • anonymous
\[\int\limits v(t) = s(t) = -16t^2 + 40t + 500\] does this look right to you ?
anonymous
  • anonymous
"se acceleration due to gravity as –32 feet per second squared"
anonymous
  • anonymous
use *
anonymous
  • anonymous
Actually, there lots of ways to do this. There is like 3
anonymous
  • anonymous
yes that is right.
anonymous
  • anonymous
i know but i have to show my work for this one so cant use physics equations sadly :(
anonymous
  • anonymous
oh wait it is wrong. Supposed to be -32t^2-40t+500
anonymous
  • anonymous
Make the position=0 by doing quadratic formula.
anonymous
  • anonymous
use your calculator divide everything by -4 first
Empty
  • Empty
Yeah as far as physics is concerned I think this would be pretty much the best way: \[K_i + V_i = K_f + V_f\] \[\frac{1}{2} m 40^2 + m 32*500 = \frac{1}{2} m v^2\] \[\sqrt{40^2 + 2*32*500} = v\]
ganeshie8
  • ganeshie8
careful, indefinite integral is unique only upto a constant
anonymous
  • anonymous
interesting, i'm getting positive 40t because of the v(t) - v(0) part (when we added the v(0) to isolate the v(t))
ganeshie8
  • ganeshie8
\[s(t)-s(0)=\int\limits_0^t v(t) -16t^2 + 40t \] so you're right, you do get \(s(t) = -16t^2+40t+500\)
anonymous
  • anonymous
okay so now just set s(t) = 0 to find when it touches the ground right?
anonymous
  • anonymous
wait i thought s (t) = -16t^2 .... not -32t^2
ganeshie8
  • ganeshie8
physicists are overjoyed by the conservation of "anything", but i think math professors still want to see the integrals mess ;)
anonymous
  • anonymous
Answer is negative because the ball's speed while hitting the ground is going downwards.
ganeshie8
  • ganeshie8
t=0 corresponds to s=500 right ?
ganeshie8
  • ganeshie8
recall that \(s(t)\) is the displacement from \(ground\) after \(t\) seconds
anonymous
  • anonymous
\[-32t^2+40t+500=0 \] got \[t=3.376s,-4.626s\] Time should only be positive so use t=3.376s \[v(t)=-64t-40\] \[v(3.376)=-256.06 m/s\] are you sure this is right @ganeshie8
anonymous
  • anonymous
shouldn't it be \[-16t^2 +40t + 500 = 0 \]
ganeshie8
  • ganeshie8
sry im not following the discussion between you and Jdosio... but you're not allowed to use direct formulas here right ?
anonymous
  • anonymous
no only calculus
anonymous
  • anonymous
okay my only positive root for v(t) is 6.978
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=-16x%5E2+%2B+40x+%2B+500
anonymous
  • anonymous
Ya use -16t^2
anonymous
  • anonymous
so to find the velocity at that time we just plug that into the formula for v(t) right ?
anonymous
  • anonymous
which gives me around -183.3030
anonymous
  • anonymous
does that look about right to you guys?
anonymous
  • anonymous
Nope it should be -40
anonymous
  • anonymous
im using v(t) = -32t + 40 is that right?
anonymous
  • anonymous
Answer is right, but the graph is wrong. Guess you do not need the graph.
anonymous
  • anonymous
okay so then does everyone agree that the answer is -183.3030 !?
anonymous
  • anonymous
Read it wrong at the beginning lol
anonymous
  • anonymous
@Empty @ganeshie8 ?
anonymous
  • anonymous
It is correct.
anonymous
  • anonymous
okay ill stick around just incase anyone changes their mind, thanks for the help guys :D
Empty
  • Empty
I got -183.3030 by doing it with the energy considerations alone, so good job! Same answer different route is a good indicator imo.
ganeshie8
  • ganeshie8
Yes the ball touches the ground after at `t=6.978` \(v(6.978) =40-32(6.978) \approx -183.303\) you want to avoid that negative sign though speed is \(183.303\) whatever unit
Empty
  • Empty
Only reason why I think they'd want the negative sign there is because they gave -32 f/s^2 as the acceleration due to gravity, which imo is kind of awkward so I don't know what they want.
ganeshie8
  • ganeshie8
acceleration is a vector, so sign makes sense speed isn't
anonymous
  • anonymous
ill just make it explicit that the velocity is downwards
ganeshie8
  • ganeshie8
they are not asking velocity
Empty
  • Empty
Yeah you're right @ganeshie8 good point.
ganeshie8
  • ganeshie8
just persisting because, when you use the terms correctly, your professor knows that you understood the concepts.. that is for @Jdosio :)
anonymous
  • anonymous
ill take note of that :)
anonymous
  • anonymous
okay something weird has happened
anonymous
  • anonymous
i found the speed of the ball at the time that the ball is at a height of 500 (the height of the cliff) and got -40 does that look oddly round to you guys ?
anonymous
  • anonymous
@ganeshie8 @Empty @Shalante ?
Empty
  • Empty
Well this is actually completely expected that the speed is the same for the same height, this is because it has no more potential or kinetic energy at this same height when you threw it. This is due to conservation of energy, which I realize is not exactly in the scope of what you're learning however it's clear at a glance to me because of this.
anonymous
  • anonymous
okay so i should stick with my 183.3030 right?

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