Physics problem pissing me off

[QueHuong]

A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 30 degrees from horizontal with a velocity of 40m/s. How far will it travel in the air?

The answer is 141 meters. But I don't understand why.
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Here is how I worked it out:

Using the following equation:
distance = (velocity initial)(time) + .5(acceleration)(time)^2

Information I gathered from the problem:
Velocity initial for the horizontal is (40)(cos30 degrees) and for the vertical it is (40)(sin30 degrees).
Horizontal acceleration is 0; vertical acceleration is 9.8m/s^2, which is gravity.

Then the next thing I have to do is find time, then use that to find the distance. Problem is that I don't know how to solve for time. I'm sure you math wizzes know how to do this.

 

[Vincent]

Figure out how long it stays in the air. This will give you the time.

 

[WouldYoURelax]

First post for me. I just finished my Physics final today and I'm feeling happy.

I'm not sure about the formula you gave. It probably works, but I think it'd be easier to understand if you break down the problem.

Since you already have the horizontal and vertical components of velocity...

You can find the time that the ball is in the air with the vertical velocity and gravity (acceleration). I THINK it's t = v(y) / g for the time it takes the ball to reach its peak. Multiply by 2 to get the total time it's in the air.

Once you have the time, you can easily find the distance it travels because the horizontal velocity is constant (no acceleration). that would be x = v(x) * t.

Hope this helps!

 

[SmackdownHotel]

Ok, easy. You only need one equation, and that is the one for range.


range = [2(V^2)/9.81] x (cos30)(sin30)


= ~141 m



There is no need for multiple steps.

 

[QueHuong]

<< x = v(x) * t >>



wouldyourelax, is there a different form of that equation that would be familiar to a high school physics student? That equation looks familiar but I can't remember what it's derived from.

 

[QueHuong]

wouldyourelax, I'm using your method but it doesn't seem to be working..unless I'm doing something wrong.


t = ((40sin30) / 9.8) * 2 = 4.08

d = (40cos30)(4.08) + (.5 * 9.8)(4.08)^2 = 222.9

 

[WouldYoURelax]

I wrote v(x) meaning the horizontal (x-axis) component of velocity. Sorry, couldn't figure out how to get that subscript in, hope it doesn't confuse you into thinking that the horizontal velocity is not constant.

I tried the problem, and I got 141m too, so it should work!

 

[WouldYoURelax]

d = (40cos30)(4.08) = 141m

Yeah you could just plug in values into the formula for the range as well. I think you're better off understanding steps than just memorizing formulas in the long run though (plus I couldn't think of the range formula off the top of my head ).

 

[Krakerjak]

<< wouldyourelax, I'm using your method but it doesn't seem to be working..unless I'm doing something wrong.


t = ((40sin30) / 9.8) * 2 = 4.08

d = (40cos30)(4.08) + (.5 * 9.8)(4.08)^2 = 222.9 >>



What you are doing is adding the second part of the equation d= vi t + .5At^2
Which involves an acceleration (due to gravity) you are finding the distance travelled (x direction) so the second half of the equation becomes zero and you are left with the first part

d = 40cos30(4.08) = 141m