# Position of object above the ground.

1. Sep 10, 2006

### Jacobpm64

The height of an object above the ground at time t is given by

s = v0t - (g/2)t2

where v0 is the initial velocity and g is the acceleration due to gravity.

(a) At what height is the object initially?
(b) How long is the object in the air before it hits the ground?
(c) When will the object reach its maximum height?
(d) What is that maximum height?

Here I go...
(a)
Just plugged in a zero for time, and solved for s to see the position of the object at time 0 (initial time).
s = v0t - (g/2)t2
s = v0(0) - (g/2)(0)2
s = 0
It's initially on the ground. (at height 0.)

(b)
I knew that the position had to be s = 0 for it to hit the ground, and I know it started from the ground, so one t-intercept would be the starting time, and the other would be the ending time, so i solved the equation for t, using s = 0.
0 = v0t - (g/2)t2
0 = t(v0 - (g/2)t)
t = 0 <--- the starting time, and i'll just worry about the other root..
v0 - (g/2)t = 0
v0 = (g/2)t
(2/g)v0 = t
t = (2v0)/g
The object is in the air (2v0)/g before it hits the ground.

(c)
For this one, since I know it's a parabola, I know that it will have a maximum where the t-coordinate will be the time of the highest point, and the s-coordinate will be the highest point. I knew that to find the vertex, the formula is -b/(2a).
s = (-g/2)t2 + v0t
Vertex = -b/(2a)
-v0 / (2*(-g/2))
-v0 / -g
v0/g
The object will reach its maximum height at time v0/g

(d)
Since I already had the time, I just plugged in the time to get the actual height, s.
s = v0(v0/g) - (g/2)(v0/g)2
s = v02/g - v02/(2g)
s = (2v02)/(2g) - v02/(2g)
s = (2v02 - v0) / (2g)
s = (v0(2v0-1)) / (2g) <--- maximum height

It was a pretty lengthy problem, and there weren't really any numbers, so I just wanted to make sure that everything was correct, thanks.

2. Sep 10, 2006

### Staff: Mentor

Looks basically correct, but it's a little weird how they ask you in (a) for the initial height. Usually you would be given an initial height as one of the initial conditions for a problem like this. But whatever.

3. Sep 10, 2006

### Jacobpm64

yeah, it'd usually be tacked on to the end of the position equation.. + 30 to mean it started 30 feet up.. I don't know.. if you've taken in physics at all, it's easy to see right off that it started from displacement 0.. *shrug* it is kind of weird, maybe they expected you not to know any physics (this is a problem in the introductory review chapter of a calculus 1 book)

that might explain it.

4. Sep 18, 2008

### lungshot

On your 4th step of part d), I think you made a simple mistake when subtracting the two fractions. The answer should simply be... ((v_0)^2)/(2g).