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anonymous
 one year ago
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.
anonymous
 one year ago
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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 @nincompoop @pooja195 @Preetha any ideas?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I believe there are two ways to do this 1) Calculate the trajectory 2) Calculate the energy change

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1presuming the ball lands on the ground and not on the cliff... even though it was thrown vertically upwards

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is a calc I problem by the way

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1maybe 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
 one year ago
Best ResponseYou've already chosen the best response.0okay s v(0) = 40 but how do we find v(t) ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Continue further, \[v(t) = \frac{d x(t)}{dt}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait slow down , how did you find s(t) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align}v(t)v(0)&=\int\limits_0^t a(t)\,dt \\~\\ &=\int\limits_0^t 32\,dt \end{align}\] you're given that \(a(t)=32\), see if you can integrate above now..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is position. 500ft cliff above the ground=+500 Thrown upward with speed of 40 feet per secondit 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
 one year ago
Best ResponseYou've already chosen the best response.0v(t)  40 = (32t^2)/2 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Position 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
 one year ago
Best ResponseYou've already chosen the best response.1nope, whats antiderivative of constant ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well you add a variable in this case t

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh woops i see what i did wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0v(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
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits v(t) = s(t) = 16t^2 + 40t + 500\] does this look right to you ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"se acceleration due to gravity as –32 feet per second squared"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually, there lots of ways to do this. There is like 3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know but i have to show my work for this one so cant use physics equations sadly :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh wait it is wrong. Supposed to be 32t^240t+500

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Make the position=0 by doing quadratic formula.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0use your calculator divide everything by 4 first

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.1careful, indefinite integral is unique only upto a constant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0interesting, 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
 one year ago
Best ResponseYou've already chosen the best response.1\[s(t)s(0)=\int\limits_0^t v(t) 16t^2 + 40t \] so you're right, you do get \(s(t) = 16t^2+40t+500\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so now just set s(t) = 0 to find when it touches the ground right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait i thought s (t) = 16t^2 .... not 32t^2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1physicists are overjoyed by the conservation of "anything", but i think math professors still want to see the integrals mess ;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Answer is negative because the ball's speed while hitting the ground is going downwards.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1t=0 corresponds to s=500 right ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1recall that \(s(t)\) is the displacement from \(ground\) after \(t\) seconds

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[32t^2+40t+500=0 \] got \[t=3.376s,4.626s\] Time should only be positive so use t=3.376s \[v(t)=64t40\] \[v(3.376)=256.06 m/s\] are you sure this is right @ganeshie8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0shouldn't it be \[16t^2 +40t + 500 = 0 \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1sry im not following the discussion between you and Jdosio... but you're not allowed to use direct formulas here right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay my only positive root for v(t) is 6.978

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=16x%5E2+%2B+40x+%2B+500

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so to find the velocity at that time we just plug that into the formula for v(t) right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which gives me around 183.3030

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0does that look about right to you guys?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope it should be 40

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im using v(t) = 32t + 40 is that right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Answer is right, but the graph is wrong. Guess you do not need the graph.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so then does everyone agree that the answer is 183.3030 !?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Read it wrong at the beginning lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay ill stick around just incase anyone changes their mind, thanks for the help guys :D

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I got 183.3030 by doing it with the energy considerations alone, so good job! Same answer different route is a good indicator imo.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Yes the ball touches the ground after at `t=6.978` \(v(6.978) =4032(6.978) \approx 183.303\) you want to avoid that negative sign though speed is \(183.303\) whatever unit

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Only 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
 one year ago
Best ResponseYou've already chosen the best response.1acceleration is a vector, so sign makes sense speed isn't

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ill just make it explicit that the velocity is downwards

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1they are not asking velocity

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah you're right @ganeshie8 good point.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1just persisting because, when you use the terms correctly, your professor knows that you understood the concepts.. that is for @Jdosio :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ill take note of that :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay something weird has happened

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 @Empty @Shalante ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Well 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
 one year ago
Best ResponseYou've already chosen the best response.0okay so i should stick with my 183.3030 right?
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