# Calculus problem with height and velocity

1. Jun 21, 2008

### Beeorz

1. The problem statement, all variables and given/known data

A stone is tossed vertically upward with an initial velocity of 23 ft/s from the top of a 15 ft building.

(a) What is the height of the stone after 0.12 s?
(b) Find the velocity of the stone after 3 s.
(c) When does the stone hit the ground?

2. Relevant equations

s(t) = s0 + v0t - (1/2)g*t^2
v(t) = v0 - g*t

3. The attempt at a solution

(a) s(t) = 15 + (23)(0.12) - (1/2)(32)(0.12^2) = 17.5296
(b) v(t) = 23 - 32(3) = -73
(c) 0 = 15 + 23t - (1/2)(32)t^2
0 = 16t^2 - 23t -15

Parts (a) and (b) are correct. I am only have problems with part (c). Whenever I solve this my answer comes up incorrect. Either my math is wrong is I'm simply not working it correctly. Any help is appreciated.

2. Jun 21, 2008

### Hootenanny

Staff Emeritus
Welcome to PF Beeorz,

You have made one 'small' error:
What is the final displacement of the stone?

3. Jun 21, 2008

### ace123

Where is the calculus? Isn't this just physics

4. Jun 21, 2008

### Beeorz

Involves rates of change and differentiation. For example, calculating the maximum height is attained when v(t) = 0. And since v(t) = s'(t), the object reaches its maximum height when the tangent line to the graph of s(t) is horizontal. If that isn't a good enough answer to your superb concern and helpfulness to the question at hand then I'd say that the problem is straight out of a calculus textbook...

On the other hand, I seemed to figure it out. It turns out that I was doing it correctly but didn't even try using the quadratic formula on [15 + 23t - (1/2)(32)t^2]. It didn't even cross my mind that this formula was necessary since I haven't even used it in years. The answer I end up with is approximately equal to 1.9246. If there is another (easier) way to solve part (c) rather than using the quadratic formula then by all means, please inform me. Thank again.