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ut4ever8
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This is my first post here so I want to start off introducing myself. My name is Ian and I'm a working on a degree in athletic training. I've done pretty well in any physics related classes so far and for the most part I think I'm hanging in there this semester, but I have come across a problem on some homework that has me stumped. Ill be the first to admit that I'm not a huge fan of physics but I am working very hard and paying a lot of money for my education and I want to get as much out of that education as I can. I don't want anyone to think I'm just looking for answers to homework, I really want to understand this stuff. So I'm hoping I can make some friends around here who can help me out and I'll do my best to help others as well. Unfortunately the only areas I'll really be helpful in are the areas that I currently teach. Sports and guitar. So if anyone needs to know how to throw a baseball or play a chord, I'd be glad to help. Anyways, on to the problem.
A homerun is hit such a way that the baseball just clears a call 26 m high located 123 m from home plate. The ball is hit at an angle of 38 degrees to the horizontal, and air resistance is negligible. Assume the ball is hit from a height 2 m above the ground. The acceleration of gravity is 9.8 m/s^2
What is the initial speed of the ball?
How much time does it take for the ball to reach the wall?
What is the speed of the ball when it hits the wall?
None are specifically given, but I have been trying to use these:
Displacement of X=Vxt + 1/2at^2 as well as the same for the displacement of Y, with Vy instead of Vx
I have been working on this for a day or so and haven't seen any success. I started with the equation for the displacement of Y since it is known as 24m. I also arranged time in that equation to become DispX/Vcos(38). After a bit of algebra to get Vo^2 on one side of the equation, I came up with {1/2(9.8ms2(123m)/cos(38)^2)}/{(123m)(tan38)-24m}. I worked that out to be Vo^2=13.48 and Vo = 3.67
I appreciate any help you guys can give me. Thank you so much for even taking the time to read this over.
Homework Statement
A homerun is hit such a way that the baseball just clears a call 26 m high located 123 m from home plate. The ball is hit at an angle of 38 degrees to the horizontal, and air resistance is negligible. Assume the ball is hit from a height 2 m above the ground. The acceleration of gravity is 9.8 m/s^2
What is the initial speed of the ball?
How much time does it take for the ball to reach the wall?
What is the speed of the ball when it hits the wall?
Homework Equations
None are specifically given, but I have been trying to use these:
Displacement of X=Vxt + 1/2at^2 as well as the same for the displacement of Y, with Vy instead of Vx
The Attempt at a Solution
I have been working on this for a day or so and haven't seen any success. I started with the equation for the displacement of Y since it is known as 24m. I also arranged time in that equation to become DispX/Vcos(38). After a bit of algebra to get Vo^2 on one side of the equation, I came up with {1/2(9.8ms2(123m)/cos(38)^2)}/{(123m)(tan38)-24m}. I worked that out to be Vo^2=13.48 and Vo = 3.67
I appreciate any help you guys can give me. Thank you so much for even taking the time to read this over.