March 12th, 2018, 02:40 AM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,952 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: 18,962 Thanks: 1606 
What initial velocity was given?

March 12th, 2018, 05:25 AM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 379 Thanks: 205 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 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,952 Thanks: 132 Math Focus: Trigonometry  Oh, sorry. I forgot. It's 10 m/s. 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: 2,751 Thanks: 1401 
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. 
March 12th, 2018, 08:27 PM  #6 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,952 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: 18,962 Thanks: 1606 
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  
Math Team Joined: Jul 2011 From: Texas Posts: 2,751 Thanks: 1401  Quote:
the specific velocity at $t = \sqrt{3} \, sec$ after launch is $10 \, m/s$ at $\theta = 60^\circ$ below the horizontal.  
March 13th, 2018, 03:54 PM  #9 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,952 Thanks: 132 Math Focus: Trigonometry 
Thank you, skeeter! I think I can solve the question now!


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