For number 19 in the UTexas homework, I was able to find the vertical and horizontal velocities of the ball but im confused on how to find the final answer of by how much the ball will clear/miss the crossbar. Anybody know how to set this up?
So first you'll want to find how long it takes to reach the field goal. Use Ax=A*cos(Theta) to get the horizontal component of the velocity. Plug that and the distance to the field goal into a formula of your choice (x= formula or v=d/t). You should then be able to find the time
Now we need to find the y-height of the football when it crosses the crossbar. As far as I know, the only formula that would work is the x= formula. So plug your numbers (remember Ay=A*sin(Theta) is the initial vertical velocity) into the formula to find the x variable (final displacement).
Find the difference between the height of the football at that moment and the crossbar and you should find how much it cleared or fell short of the bar.
well i didnt know how to start a new blog so im just commenting on this one..im really confused with number 18 and i dont even know where to begin..also i still dont understand 19 even after that exaplaination :/
Back to what Emily asked: since there is no horizontal acceleration, the average horizontal velocity equals the initial horizontal and the final horizontal velocity
Let's make a plan on how we're going to solve this. What do we need? The height of the football when it reaches the field goal. How do we get the right value for time? Find how long it takes for the football to cross the field.
Alright, first step: find the time it takes for the football to reach the field goal. We have the angular velocity, but we need the horizontal because right now we're just worrying about the horizontal stuff. The formula for that is Ax=A*cos(Theta). A is the velocity the question gives us and Theta is the angle the question gives us. Find Ax. Now we need to find the time. Let's use the v=d/t formula (because there is no horizontal acceleration, the velocity is constant, the average velocity is the same as the initial and final). Plug in the values and find t. Now we can use that for the next step.
Finding the height: Easiest way is to use the x=xo+vot+1/2at^2 formula. Remember, we're finding the height so we're going to use the vertical variables. I said up was positive. Then xo was 0, a was -9.8, t was the answer I found earlier, and vo is the vertical component of the velocity (Ay = A*sin(Theta)). Find x to find how high the football is.
Getting the answer: Just find the difference between the the height of the football and the crossbar. That should do it.
since horizontal acceleration is always 0 there is no chance in velocity, therefore the average velocity is the same thing as the initial or final acceleration.
11 comments:
So first you'll want to find how long it takes to reach the field goal. Use Ax=A*cos(Theta) to get the horizontal component of the velocity. Plug that and the distance to the field goal into a formula of your choice (x= formula or v=d/t). You should then be able to find the time
Now we need to find the y-height of the football when it crosses the crossbar. As far as I know, the only formula that would work is the x= formula. So plug your numbers (remember Ay=A*sin(Theta) is the initial vertical velocity) into the formula to find the x variable (final displacement).
Find the difference between the height of the football at that moment and the crossbar and you should find how much it cleared or fell short of the bar.
when im finding the horizonital component of the velocity, im multiplying cos47degrees by 21m/s right?
yeah its 21cos(47). then use that velocity in the v=d/t formula to get the time.
But wouldnt that equation not work because thats for average velocity? Thanks for the help btw
well i didnt know how to start a new blog so im just commenting on this one..im really confused with number 18 and i dont even know where to begin..also i still dont understand 19 even after that exaplaination :/
Back to what Emily asked: since there is no horizontal acceleration, the average horizontal velocity equals the initial horizontal and the final horizontal velocity
so then the initial vertical velocity would be zero right?
Let's make a plan on how we're going to solve this. What do we need? The height of the football when it reaches the field goal. How do we get the right value for time? Find how long it takes for the football to cross the field.
Alright, first step: find the time it takes for the football to reach the field goal. We have the angular velocity, but we need the horizontal because right now we're just worrying about the horizontal stuff. The formula for that is Ax=A*cos(Theta). A is the velocity the question gives us and Theta is the angle the question gives us. Find Ax. Now we need to find the time. Let's use the v=d/t formula (because there is no horizontal acceleration, the velocity is constant, the average velocity is the same as the initial and final). Plug in the values and find t. Now we can use that for the next step.
Finding the height: Easiest way is to use the x=xo+vot+1/2at^2 formula. Remember, we're finding the height so we're going to use the vertical variables. I said up was positive. Then xo was 0, a was -9.8, t was the answer I found earlier, and vo is the vertical component of the velocity (Ay = A*sin(Theta)). Find x to find how high the football is.
Getting the answer: Just find the difference between the the height of the football and the crossbar. That should do it.
No, the initial vertical velocity is not zero. The initial vertical velocity is A*sin(Theta). In your case that would be 21*sin47 I believe.
since horizontal acceleration is always 0 there is no chance in velocity, therefore the average velocity is the same thing as the initial or final acceleration.
Just felt the need to clarify.
thanks so much!
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