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November 26, 2014

November 26, 2014

Posted by **mark** on Monday, October 3, 2011 at 12:11am.

- physics -
**mardin**, Sunday, November 25, 2012 at 4:03pm1.) First, solve in the y direction. You know the ball starts at 2m and ends at 3.05m. You also know that acceleration is -9.8m/s^2, and that initial velocity in the y direction is going to be the overall initial velocity times sin45. You then use the equation x-x0=v0t+.5at^2, which after plugging in your values is 1.05=v0sin(45)t-4.9t^2. Now you just need something to plug in for t, and to do this, create an equation in the x direction.

2.) In the x direction, acceleration is going to be zero, since the problem does not tell you about any air resistance. Distance is 10m, and the initial velocity in the x direction is going to be the overall initial velocity times cos45. So using the same equation as in 1.) except plugging in numbers from the x direction, you get 10=v0cos(45)t, with no acceleration. You can then see that t=10/(v0cos45).

3.) Plug this value for t back into the equation in the y direction. 1.05=10v0sin(45)/v0cos(45)-490/(v0cos45)… This simplifies to 8.95=490/(v0cos45)^2, and solving for v0 would give you v0 = (490/(8.95cos^2(45))^(1/2).

So your final answer is v0=10.46m/s.

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