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March 12th, 2018, 02:40 AM   #1
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[ASK] Confused About Velocity

Something was thrown with an elevation angle $\displaystyle 60^{\circ}$. Determine its velocity and direction after:
a. $\displaystyle \frac{1}{2}\sqrt3$ second
b. $\displaystyle \sqrt3$ second

I tried solving the (a) question by substituting the angle to the equation $\displaystyle y=v_0\sin\alpha t-\frac{1}{2}gt^2$ and got 0 (the book uses $\displaystyle 10m/s^2$ as the gravity acceleration), which is impossible since at speed 0 the thing must have been stopped, but when I determined the time when it reached the maximum distance using the formula $\displaystyle t_{\text{x max}}=\frac{2v_0\sin\alpha}{g}$ I got $\displaystyle \sqrt3$ second which means the thing hasn't stopped yet in $\displaystyle \frac{1}{2}\sqrt3$ second. Did I make a mistake somewhere?
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March 12th, 2018, 04:02 AM   #2
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What initial velocity was given?
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March 12th, 2018, 05:25 AM   #3
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The problem with formulas is people remember only the formula and forget what it computes. $v_0 \sin \alpha t - \frac{1}{2}gt^2$ is not the formula for the velocity. It is the formula for the vertical projection of the position. Do you know the difference?

Last edited by skipjack; March 13th, 2018 at 05:08 AM.
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March 12th, 2018, 04:58 PM   #4
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Quote:
Originally Posted by skipjack View Post
What initial velocity was given?
Oh, sorry. I forgot. It's 10 m/s.

Quote:
Originally Posted by SDK View Post
The problem with formulas is people remember only the formula and forget what it computes. $v_0 \sin \alpha t - \frac{1}{2}gt^2$ is not the formula for the velocity. It is the formula for the vertical projection of the position. Do you know the difference?
No.

Last edited by skipjack; March 13th, 2018 at 05:08 AM.
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March 12th, 2018, 06:31 PM   #5
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Velocity components of a projectile acting solely under the influence of gravity ...

$v_x = v_0 \cos{\theta_0}$, a constant

$v_y = v_0 \sin{\theta_0} - g \cdot t$

To determine velocity, you need speed at time $t$ ...

$|v| = \sqrt{v_x^2 + v_y^2}$

and direction at time $t$ ...

$\theta = \arctan \left(\dfrac{v_y}{v_x}\right)$

You have ...
$v_0=10 \, m/s$, $\theta_0 = 60^\circ$, $g=9.8 \, m/s^2$, and two different times.
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March 12th, 2018, 08:27 PM   #6
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Thanks. Btw when I was trying to solve the (b) question, I got negative velocity. Does it mean that the thing has already moving downward?
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March 13th, 2018, 05:13 AM   #7
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Regarding skeeter's reply, it was mentioned that the book uses $g=10 \, m/s^2$ rather than $g=9.8 \, m/s^2$.
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March 13th, 2018, 10:28 AM   #8
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Quote:
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Thanks. Btw when I was trying to solve the (b) question, I got negative velocity. Does it mean that the thing has already moving downward?
the y-component of velocity is negative ... yes, it means the projectile is moving downward.

the specific velocity at $t = \sqrt{3} \, sec$ after launch is $10 \, m/s$ at $\theta = -60^\circ$ below the horizontal.
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March 13th, 2018, 03:54 PM   #9
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Thank you, skeeter! I think I can solve the question now!
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