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 2 years 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?
 2 years 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?

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bibby
 2 years ago
Best ResponseYou've already chosen the best response.2@zepp @AravindG @tcarroll010

richyw
 2 years ago
Best ResponseYou've already chosen the best response.2integrate 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!

bibby
 2 years ago
Best ResponseYou've already chosen the best response.2also one more thing. When it asks for "find displacement when acceleration is 0", dpo I also plug back into s(t)?

richyw
 2 years ago
Best ResponseYou've already chosen the best response.2uh, they told you to assume that acceleration is 32...

richyw
 2 years ago
Best ResponseYou've already chosen the best response.2maybe that's after it hit the ground?

zepp
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, 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.

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

bibby
 2 years ago
Best ResponseYou've already chosen the best response.2zepp: 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.

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

bibby
 2 years ago
Best ResponseYou've already chosen the best response.2I guess I meant the technique and wording and not solving. I'm sure derivatives aren't exclusive to calculus :D.

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

zepp
 2 years ago
Best ResponseYou've already chosen the best response.0^ could easily be derived using simple algebra :D

zepp
 2 years ago
Best ResponseYou've already chosen the best response.0Let'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 \(\large \Delta x=xx_0\)

zepp
 2 years ago
Best ResponseYou've already chosen the best response.0dw: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 :)

bibby
 2 years ago
Best ResponseYou've already chosen the best response.2ah sorry I closed the question. I don't think pm will work out but good luck if you guys can still type

richyw
 2 years ago
Best ResponseYou've already chosen the best response.2you 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)

zepp
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, acceleration is constant, it's the gravitational pull, usually, so :P
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