How Far Down Does the Stone Strike the Opposite Building Wall?

[chudd88]

Note: I figured this out. See my follow-up. Thanks.

Homework Statement

Two tall buildings of equal height stand with vertical walls 40m apart. A stone is thrown from the top of one at an angle of 25 degrees below the horizontal and directly at the other building with a velocity of 18 m/s. How far down from the top does it strike the opposite wall?

Homework Equations

The Attempt at a Solution

Note, first of all, that the picture it shows in my book has the stone coming off the very top of the building. There's no one throwing the stone, so it's being thrown at y=0, where the very top of the building is 0.

I've worked this problem several times, and I keep getting the same (wrong) answer, according to the answers at the end of my book. Here's my process:

25 degrees down, with an initial velocity of 18 m/s means the horizontal component of the velocity is: 18 m/s * cos(-25) = 16.31 m/s

The vertical component is 18 m/s * sin(-25) = -7.607 m/s

So far so good?

The stone will travel 40 m horizontally at 16.31 m/s, which means the time is 40 m / 16.31 m/s = 2.45s

So, with that, I can use my initial vertical velocity of -7.607 m/s, acceleration of -9.81 m/s^s, and time = 2.45s, to calculate the y.

y = (-7.607m/s)*(2.45s) + (1/2)*(-9.81m/s^2)(2.45^2) = -48.08 m

That's not the correct answer, according to my book. Can anyone spot what I'm doing wrong?

(The book says the answer is 41.8m, btw)

Thanks.

 

[chudd88]

Well, I figured it out, sort of. It turns out I believe the answer in my book is wrong. Just because I had exhausted all other possibilities I could think of, I hypothesized that maybe the book accidentally used RADIANS instead of DEGREES when calculating the sin and cos values. When I switched my calculator to RADIANS, and ran the same calculations as I stated in my original post, I got the book's answer.

So, unless someone feels using RADIANS is appropriate for this problem (it isn't AFAIK), then I'll just conclude the book's wrong, and my original answer is correct.

 

[Rake-MC]

regardless of whether degrees or radians are being used, you should come out on top with the same values for the x and y components of initial velocity.

 

[chudd88]

regardless of whether degrees or radians are being used, you should come out on top with the same values for the x and y components of initial velocity.

I don't follow. The initial x velocity is 18 m/s * cos(-25).

In radian measure, cos(-25) = 0.9912
In degree measure, cos(-25) = 0.9063

I was using degree measure in my initial post, and I've concluded that the book was (incorrectly) using radian measure. Maybe I'm wrong, but when radian measure is used, I get the book's answer. (Note, however, that they didn't use radian measure when calculating the sin(-25)).

So, from what I understand, using degrees or radians will completely change my initial x and y velocity components.

 

[Rake-MC]

Ahh I see, in that case the book is incorrect, however that's not the correct interchange of radians and degrees. If the book specified that it's 25 degrees under horizontal, and you did -25 radians and it worked out, I can't argue that. But degrees and radians are proportional by \frac{1}{180} in that:

25^o = \frac{25\pi}{180}

As I interpreted it, you had entered the RHS as your value of radians (\frac{5pi}{36})