# Throwing a ball from a building [a trajectory problem].

1. Sep 13, 2007

### niyati

A ball is thrown from the top of a building upward at an angle of 66 degrees to the horizontal and with an initial speed of 8 m/s. The ball is thrown at a height of 34 m above the ground and hit s the ground 11.3347 m from the base of the building. The acceleration of gravity is 9.8 m/s^2.

The answer (or how to do it) is basically given to us. We are suppose to use y[displacement] = v[yi](t) + .5a(t^2), with y = -34, v[yi] being 8sin66, and a = - 9.8.

The goal, being of course, to get the time it takes for the ball to follow through the entire trajectory. However, I get a positive and negative root, and I am not sure why.

t = 3.48, -1.99

What does the neg. root represent? If I made the "top" as 34, and y[final] as 0 (making my total vertical displacement as positive 34) how would I have to figure my equation and alter it to suit this new perspective?

I'm not done, though.

With the positive time (should these previous calculations be correct), I can multiply that by -9.8 m/s^2, and then subtract that from 8sin66, or about 7.31 m/s to get the final vertical velocity component. Since horizontal velocity is constant, 8cos66 would become my final horizontal component. Squaring both terms, adding, and then taking the sq. rt. of the sum gives me my answer.

...but yea. It's the first part that confuses me conceptually. If I understand that part, perhaps the second part will fall into place. I would appreciate any explanation. :D

Last edited: Sep 13, 2007
2. Sep 13, 2007

### mjsd

firstly, you should appreciate the fact that you are solving a quadratic equation over real numbers and so there is no surprise that you get two answers. In this situation, only one of them actually make sense. And you reject the other one based on common sense and that we seek an answer for t > 0.

for second part, I don't actually get t = 0.502 or -0.287 as answers (check again)

3. Sep 13, 2007

### niyati

Yes, I know. That is why I altered my answers a few minutes after posting. I realized my mistake for the second part.

:D

4. Sep 15, 2007

### dynamicsolo

As mjsd said, you only want the answer that satisfies t > 0 because you called
t = 0 the time at which the ball is thrown. The quadratic equation you set up,
-34 = 7.308(t) - 4.905(t^2), however, is simply the equation of a parabola with its vertex at the origin (t = 0, y = 0), for which you are seeking the values of t at which y = -34. The mathematical expression has *nothing* to do with physics: you could as easily ask when y would equal -1034 m (with the ball being a kilometer underground) and you would still find numerical solutions.

By the same token, this same equation can also describe the parabola followed by the ball had it been thrown *from street level* at a moment t = -1.99 sec., with the appropriate initial velocity, so that it reaches the top of the building at time t = 0, now moving at 8 m/sec at an angle 66 deg. above the horizontal, passes through its peak, falls back below the top of the building (note that every height except the peak is attained at two different values of t), and finally strikes the ground at t = 3.48 sec. We are free to select the moment we call t = 0; doing so in this way leads to the values you found. Because we know for this problem, though, that we only threw the ball at t = 0 from the top of the building, we then also know to ignore the value t = -1.99 sec.: *our* ball was never on that part of the mathematical parabola.

Incidentally, you can avoid this ambiguity entirely by working with the horizontal velocity instead. Since we are presumably neglecting air resistance, there are no horizontal forces on the ball, so it travels at a constant horizontal speed. That speed is
vx = 8 cos 66 = 3.254 m/sec. We are told that the ball traveled horizontally by 11.3347 m (we're assuming the ball was thrown from the top *edge* of the building), so its travel time is 11.3347 m / 3.254 m/sec = 3.483 seconds; the horizontal position equation is *linear*, so there is *only* this one solution.

Your question illustrates an important point about the *interpretation* of mathematical models for physics that few lecturers or textbooks think to remark upon...

The vertical displacement is final y position minus initial y position. You called vertically upward positive (so that the vertical acceleration is -9.81 m/sec^2) and the ball's starting position y = 0, so the street level is y = -34 m and the vertical displacement is (-34) - 0 = -34 m. If you call the street level y = 0 and the top of the building y = 34 m, which is still consistent with "positive upward", your displacement will still be 0 - 34 =