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An object is in motion according to the conditions:

$a(t)= \langle 0, 0,−32\rangle$, and $v(0)= \langle50, 0, 50\sqrt{3}\rangle$, and $r(0)=\langle 0, 0, 0\rangle$. Find the velocity and position functions for this motion. What is the maximum height (z value) that this object reaches?

Here is what I did:

$\overrightarrow v(t)= \int \overrightarrow a(t)dt = \int dt + \int dt + \int-32dt = C_1 + C_2 + -32t + C_3 = \langle C_1, C_2, -32t + C_3 \rangle$

So I assume now I solve for my constants. The first two are simple:

$\overrightarrow v(0) = \langle 50, 0, 50\sqrt{3}\rangle$

$C_1 = 50$, $C_2 = 0$, $C_3 = 50\sqrt{3}+32t$

$\overrightarrow v(t) = \langle 50, 0, 50\sqrt{3}+32t \rangle$

I'm not sure if $C_3$ is correct? Assuming it is, I then just take the integral of my new $\overrightarrow v(t)$, right?

$\overrightarrow r(t) = \int \overrightarrow v(t)dt = \int 50 dt + \int dt + \int 50\sqrt{3}dt = 50t+C_4 + C_5 + 50\sqrt{3}t + C_6 = \langle 50t+C_4, C_5, 50\sqrt{3}t+C_6\rangle $

Solving for my constants I get:

$\overrightarrow r(0) = \langle 0,0,0\rangle $

$C_4=-50t$, $C_5=0$, $C_6 = -50\sqrt{3}$

$\overrightarrow r(t) = \langle -50t, 0, -50\sqrt{3}\rangle$

Is this method correct? If it is, how do I find the maximum height the z value reaches?

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  • $\begingroup$ Very importan thing left out: What forces are present? Air friction, gravity? $\endgroup$ – zickens Nov 15 '15 at 3:35
  • $\begingroup$ @JohnDouma I did that because $50t+C_4=0$ so $C_4=-50t$ and I used the same method for $C_6$. Was this incorrect? $\endgroup$ – hax0r_n_code Nov 15 '15 at 3:35
  • $\begingroup$ The signs are reversed in $x$ and $z$ and $z$ should be $50\sqrt{3}t+16t^2$. $\endgroup$ – John Douma Nov 15 '15 at 3:35
  • $\begingroup$ @zickens haha I'm assuming this is taking place in a vacuum. Like everything in the first two semesters of physics :-P $\endgroup$ – hax0r_n_code Nov 15 '15 at 3:37
  • $\begingroup$ Notice you made a mistake when calculating $C_{3}$ for $\vec{v}$; it should be $-32$. $\endgroup$ – zickens Nov 15 '15 at 3:37
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You have three components. The $x$, $y$ and $z$.

For $x$ we get a constant velocity because there is no acceleration so $$v_x=50\implies \frac{dx}{dt}=50\implies x=50t+C$$

But we know that $x(0) = 0$ so $50(0)+C=0\implies C=0\implies x= 50t$ $$\therefore x=50t$$

For $y$ we also get a constant velocity and $$v_y=0\implies \frac{dy}{dt}=0\implies y=C$$ and since $y(0)=0$, $C=0$ so $$y=0$$

The $z$ direction defines a negative acceleration so we get $$a=-32\implies \frac{dv}{dt}=-32\implies v=-32t+C$$ and since $v_y(0)=50\sqrt{3}$ we get $$-32(0)+C=50\sqrt{3}\implies C=50\sqrt{3}$$ $$\therefore v_z=-32t+50\sqrt{3}t$$

Finally, $$\frac{dz}{dt}= -32t+50\sqrt{3}t\implies z=-\frac{1}{2}(32)t^2+50\sqrt{3}t+C=-16t^2+50\sqrt{3}t + C$$

and since $z(0)=0$ we get that $C=0$ so $$z=-16t^2+50\sqrt{3}t$$

The object reaches its highest height when $v_z=0\implies -32t+50\sqrt{3}=0\implies t=\frac{25\sqrt{3}}{16}$

Plug that into your expression for $z$ to get $z\approx 117.2$

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  • $\begingroup$ Thanks, John! Did you see my comment with regards the the reversed signs? Does that make sense or did I do it incorrectly? $\endgroup$ – hax0r_n_code Nov 15 '15 at 3:53
  • $\begingroup$ @inquisitor I don't see it. Why were the signs reversed? Did you change coordinates? $\endgroup$ – John Douma Nov 15 '15 at 3:55
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    $\begingroup$ @inquisitor I see your explanation. No. That is incorrect. $50t + C_4\ne 0$. You are using initial conditions so $50(0) + C_4=0\implies C_4=0$. $\endgroup$ – John Douma Nov 15 '15 at 3:57
  • $\begingroup$ So the terms I integrated should be set equal to my initial values, in this example $0$? $\endgroup$ – hax0r_n_code Nov 15 '15 at 4:03
  • $\begingroup$ @inquisitor Yes. Let me fill in the details. $\endgroup$ – John Douma Nov 15 '15 at 4:04

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