Physics Problem Needing Help?

Question

Please help!!! I am SO lost!!!

A placekicker must kick a football from a point 36.0 m (about 39.0 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a velocity of 20.3 m/s at an angle of 54.0° to the horizontal.

A) By how much does the ball clear OR fall short of clearing the crossbar (in m)?

B) Does the ball approach the crossbar while still rising or while falling?

Answer 1

I don't like doing hw for other people, but I can give you some hints (even though this will likely cost me the vote for best answer. Shucks! Getting that recognition is so important to me!)

a.
You have a velocity and an angle relative to the ground. Thus you can define the direction parallel with the ground as the x-axis and the direction to the sky as the y-axis. Using a bit of trig, you can figure out the x and y components of the given velocity (make a right triangle with the given velocity as a hypotenuse and pointing 54 degrees off the ground.

You know have two things to analyze: The stuff happening in the x-direction, and the stuff happening in the y-direction.

In the x direction, no forces act on the ball (well, technically, there's air resistance, but that's beyond the scope of this problem, I assume?), so you therefore have no acceleration and therefore a constant velocity.

Under constant velocities, distance is simply velocity times time. You figured out the horizontal velocity earlier, so using that and the given distance, you can figure out how long it takes for hte ball to get to the goal.

That was the easy part....

You now have to look at the y-component, to know how high the ball is when it reaches the goal. The y direction is governed by a few equations, e.g. y^2=y_0^2+v_0*t+(1/2)a*t^2. You know a (usually just -9.8m/s^2), and you got t. y_0 is zero, so you can use it to solve for y at the t you solved earlier. Is that y higher than the crossbar? If so, you made it. If not, you didn't.

b.
This part is kind of tricky. Re-using y^2=y_0^2+v_0*t+(1/2)a*t^2, you'll get it.

It's like this: The ball moves parabolically, and will be at each particular height twice (once on its way up and again back down). For example, it's height will go something like 0 feet, 1 foot, 2 feet, 3 feet, 2 feet, 1 foot, 0 feet. Get it? So, what you can do is, plug in the y you solved for in part A and solve for t. You should get two t's as answers. The first one is when the ball is rising, and the second is when the ball is falling. Which one matches the t you got at the beginning?

IN SUMMARY: This sort of problem (a type of 2-D kinematics) boils down to two things usually. First, look in the x direction to get some sort of time. Then use that time to solve for things in the y direction. That's pretty much it.

I know this is a lot to digest in one shot, so just go slowly, and write down each step. Make sure you understand why each step is the way it is. Good luck, and don't get discouraged. Physics is fun! It's frustrating at first, but as you go on, you'll feel yourself getting smarter, and one day, everything will suddenly light up.

Answer 2

What are they doing to our kids in schools these days? Who CARES! What does it accomplish for you in the future? Is this a joke? You think YOU are lost? I can spell any word in the English/American language, and can use words I hardly know in a sentence. But MATH???? WHY??? Why do they torture us?

Answer 3

I dont know for sure but try reading the question slow and see what makes you feel you picked the right answer. Trust me it works.

Answer 4

since the crate is moving at a constant speed the frictional force is the same as the force needed to keep it at that speed