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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: 21,028 Thanks: 2259 What initial velocity was given? March 12th, 2018, 05:25 AM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 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: 3,034 Thanks: 1621 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: 21,028 Thanks: 2259 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.
Attached Images velocity_projectile.jpg (23.8 KB, 0 views) 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! Tags confused, velocity Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post IzzyD Physics 3 June 3rd, 2017 12:29 PM quarkz Calculus 0 April 18th, 2014 05:34 AM Josh Algebra 2 November 13th, 2013 01:38 PM Debjani Algebra 1 April 24th, 2009 09:57 AM namarsha Real Analysis 2 March 2nd, 2009 03:52 PM

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