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(a) A rocket is fired straight up from a tower 110 ft above the ground with an initial velocity of 224 ft/s. (Assume the acceleration due to gravity is −32 ft/s2).
When does the rocket reach its maximum height? and at this time, how far is the rocket above the ground?
 one year ago
 one year ago
(a) A rocket is fired straight up from a tower 110 ft above the ground with an initial velocity of 224 ft/s. (Assume the acceleration due to gravity is −32 ft/s2). When does the rocket reach its maximum height? and at this time, how far is the rocket above the ground?
 one year ago
 one year ago

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bibbyBest ResponseYou've already chosen the best response.1
@zepp @AravindG @tcarroll010
 one year ago

richywBest ResponseYou've already chosen the best response.2
integrate the acceleration with respect to time to get the velocity, the constant of integration is the initial velocity. then integrate this velocity function with respect to time, the constant of integration will be the initial height. then you have the position as a function of time. to figure out the maximum height you just need to solve for when v(t)=0, then plug that t value into the position function to get the height!
 one year ago

bibbyBest ResponseYou've already chosen the best response.1
also one more thing. When it asks for "find displacement when acceleration is 0", dpo I also plug back into s(t)?
 one year ago

richywBest ResponseYou've already chosen the best response.2
uh, they told you to assume that acceleration is 32...
 one year ago

richywBest ResponseYou've already chosen the best response.2
maybe that's after it hit the ground?
 one year ago

zeppBest ResponseYou've already chosen the best response.0
Okay, this is a physics question, but whatever. Given the initial height, let's call it \(\large y_0\), 110 ft; We are looking for the final height, \(\large y\). We also know that when the rocket will reach its highest point, it's velocity is 0, \(\large v=0\), and the acceleration would be \(\large 32ft/s^2\). From the freefall equations of kinematics, using the formula: \(\large v^2=v_0^2+2a(yy_0)\) So, \(\large 0=(244ft/s)^22a(y110ft)\\ \large 59536=2a(y110)\\\large 930.25=y110\\\large y=930.25+110=1040.25\) The final height would be 1040.25 feet in the sky.
 one year ago

zeppBest ResponseYou've already chosen the best response.0
If you want to find \(t\), then plug everything in \(\large y=y_0+\frac{1}{2}(v_0+v)t\) and solve for \(t\).
 one year ago

bibbyBest ResponseYou've already chosen the best response.1
zepp: that's the thing. It's one of my previous calc finals and I've never seen it worded like this although I'm pretty sure integration is how to go about it.
 one year ago

zeppBest ResponseYou've already chosen the best response.0
Of course, it would be solved using integration, but that's a classical mechanics introductory course question ;x
 one year ago

bibbyBest ResponseYou've already chosen the best response.1
I guess I meant the technique and wording and not solving. I'm sure derivatives aren't exclusive to calculus :D.
 one year ago

richywBest ResponseYou've already chosen the best response.2
\[a=\frac{dv}{dt}\]\[\int a\,dt=\int dv\]\[v=at+v_0\]\[v=\frac{dx}{dt}\]\[\int (at+v_0)dt=\int dx\]\[x=\frac{1}{2}at^2+v_0t+x_0\]
 one year ago

zeppBest ResponseYou've already chosen the best response.0
^ could easily be derived using simple algebra :D
 one year ago

zeppBest ResponseYou've already chosen the best response.0
Let's go from the acceleration, which is \[\large a=\frac{\Delta v}{\Delta t}=\frac{v_fv_i}{t_ft_i}\]If we set \(t_i=0\) and \(t_f=t\), we get \(\large a=\frac{vv_0}{t}\), or \(\large v=v_0+at\) Now the formula to find the displacement is \(\large \Delta x=xx_0\)
 one year ago

zeppBest ResponseYou've already chosen the best response.0
dw:1355785810205:dw We know that the displacement could be given by \(\large \Delta x = v_{av}\Delta t=\frac{(v_f  v_i)}{2}\) So the distance would be \(\large x= x_0 + \frac{1}{2}(v_0+v)t\) Now, if you substitute \(v=v_0+at\) into the formula above, you'll get \(\large x = x_0+v_0t+\frac{1}{2}at^2\) by algebraic manipulations :)
 one year ago

bibbyBest ResponseYou've already chosen the best response.1
ah sorry I closed the question. I don't think pm will work out but good luck if you guys can still type
 one year ago

richywBest ResponseYou've already chosen the best response.2
you can type after it's closed anyways! and yeah but this works because acceleration is constant right? personally i'd just do it the faster way! (i can never memorize the kinematic eqns though)
 one year ago

zeppBest ResponseYou've already chosen the best response.0
Yes, acceleration is constant, it's the gravitational pull, usually, so :P
 one year ago
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