HELP PLEASE (Physics and Math inclined)

<p>Please solve this problem with steps… I will be forever grateful!</p>

<li>A pitched ball is hit by a batter at a 45 degree angle and just clears the outifled fence 98 meters away. Assume that the fence is at the same height as the pitch and find the veloicty when it left the bat. </li>
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<p>It is dealing with Motion in two dimensions (thts the chapter I am up to)
iTS worth like 15 points extra credit - so please help if you know!</p>

<p>Help with homework!! I don't know if thats legal. I think it goes like this: (correct anyone me if I'm wrong).</p>

<p>Since the ball goes off at a 45º angle, both Vx and Vy must be the same, right? Then we say that the peak of the ball's flight will be at the 49m mark, and when it peaks, Vy=0. (Gravity has been slowing it down up till then, and after the peak, it starts to pull it out of the sky). Then we solve this equation: V = Vo + at. Well, V = 0 (as above) and a = -9.8, so we get Vo=9.8t. All that is in the y dimension. Then in the x dimension we solve x = Vot + .5at^2 Acceleration for x is zero (we ignore air resistance) and the distance we want to calculate for is to the half way mark, 49m, where Vy = 0. So we get 49 = Vot. Now we substitute our other Vo equation into this one, and get 49 = 9.8t^2, and that means that the time to the half way mark is 2.24seconds. So double the time is 4.48 seconds, and that means the ball has to travel in the x dimension at 21.9m/sec to travel 98m in 4.48sec. Well, thats just Vx. So we divide by cos(45º) to get the total velocity at the 45º angle, and get about 31m/sec.</p>

<p>Instead of all that you could have used the range equation, which is much easier and gives the same result: R = [(Vo)^2 sin(2@)]/g where @ is theta.</p>

<p>Schoolwork questions are best answered here:
Homework</a>, Coursework, & Textbook Questions
You need to show some efforts though.</p>

<p>Yeah, physicsforums is great</p>