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 March 12th, 2018, 02:40 AM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry [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?
 March 12th, 2018, 04:02 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,819 Thanks: 2158 What initial velocity was given?
 March 12th, 2018, 05:25 AM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics 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.
March 12th, 2018, 04:58 PM   #4
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Quote:
 Originally Posted by skipjack What initial velocity was given?
Oh, sorry. I forgot. It's 10 m/s.

Quote:
 Originally Posted by SDK 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.

 March 12th, 2018, 06:31 PM #5 Math Team     Joined: Jul 2011 From: Texas Posts: 2,978 Thanks: 1573 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. Thanks from Monox D. I-Fly
 March 12th, 2018, 08:27 PM #6 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry 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?
 March 13th, 2018, 05:13 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,819 Thanks: 2158 Regarding skeeter's reply, it was mentioned that the book uses $g=10 \, m/s^2$ rather than $g=9.8 \, m/s^2$.
March 13th, 2018, 10:28 AM   #8
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Quote:
 Originally Posted by Monox D. I-Fly 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 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Thank you, skeeter! I think I can solve the question now!

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