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January 20th, 2017, 06:59 AM  #1  
Member Joined: Feb 2014 Posts: 91 Thanks: 1  Could someone give me a hint on how to get the initial velocity
I'm working on some homework. I need to find the initial velocity. The question I'm working on is stated below. After some google research and Khan academy I thought that this was easy so I substituted 0 for time t and came to the results of 50m. I then tried to solve for acceleration and that came out to be zero. These solution just don't feel right to me like I did something wrong, it my gut feeling. The (initial velocity iV) should be distance divided by time right? so V is also zero despite a distance of negative 50 being covered...... Please guide me. int (Time) t = 0; int (distance) m = 50; int (acceleration) a = 0; Quote:
 
January 20th, 2017, 07:30 AM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575  Quote:
$y_0$ is the initial position relative to some designated zero position $v_0$ is the initial velocity $g$ is the magnitude of the acceleration due to gravity I have a problem with the coefficient of the quadratic term ... $20 = \dfrac{1}{2}g \implies g = 40$. On the Moon, the value of $g$ is known to be about $1.6 \, m/s^2 = 5.3 \, ft/s^2$  
January 20th, 2017, 07:46 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,835 Thanks: 2162 
I assume it was intended that y(t) = 50 + 100t  20t^2. It follows that y'(t) = 100  40t and y''(t) = 40. Hence y'(0) = 100, y(0) = 50, and y''(0) = 40. It just remains to add the appropriate units. 
January 20th, 2017, 07:59 AM  #4  
Member Joined: Feb 2014 Posts: 91 Thanks: 1  Quote:
How were you able to pick apart the equation to tell what was the initial velocity?  
January 20th, 2017, 08:08 AM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575  Quote:
$x(t) = x_0 + v_0 \cdot t + \dfrac{1}{2}at^2$ modified for motion under the influence of gravity ... $y(t) = y_0 + v_0 \cdot t  \dfrac{1}{2}gt^2$ ... so all you have to do is recognize the values of the respective coefficients and what they represent. If you are familiar with the relationship between position, velocity, and acceleration derived from the calculus, then have a look at post #3 for that explanation.  

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