Question

Let’s suppose you throw a ball straight up with an initial speed of 50 feet per second from a height of 6 feet.

Find the parametric equations that describe the motion of the ball as a function of time.
How long is the ball in the air?
Determine when the ball is at maximum height. Find its maximum height.

Answer 1

Satellite, you were on this really fast.. :P

Answer 2

usual formula is
\[g(t)=-16t^2+v_0t+h_0\] here
\[v_0=50,h_0=6\]

Answer 3

i meant \(h(t)\)

Answer 4

in any case
\[h(t)=-16t^2+50t+6\] for your example.
to see how long it is in the air, set \(h(t)=0\) and solve for \(t\)

Answer 5

I just saw how my textbook wanted me to do it, i'll show you how it wanted.

Answer 6

max height is at the vertex of this parabola, which opens down
if your text wants to you take a limit, tell it to go pound sand. no point in it
vertex of the parabola is where \(t=-\frac{b}{2a}\)

Answer 7

x=v[0]cos(theta))t+h
y=-(1/2)gt^2+(v[0]sintheta)t+h

Answer 8

http://screensnapr.com/v/FpJu8X.png

Answer 9

Well they just use different equations from you, whats the difference though?

Answer 10

ball in this case is thrown straight up, so \(\theta =90, \cos(90)=0,\sin(90)=1\)

Answer 11

I like your way better :P