A ball is thrown vertically upward with an initial speed of 80 ft/s. Its height after t seconds is given by h=80t-16t^2.
a. How high does the ball go?
b. When does the ball hit the ground?

Question

A ball is thrown vertically upward with an initial speed of 80 ft/s. Its height after t seconds is given by h=80t-16t^2.
a. How high does the ball go?
b. When does the ball hit the ground?

anonymous · Answer

1.

$$h = 80t - 16t^2$$

Evaluate t from here..

anonymous · Answer

How? @waterineyes

anonymous · Answer

I am also not getting this..

anonymous · Answer

max height is at the vertex 
put $x=-\frac{b}{2a}=-\frac{80}{2\times -16}=\frac{5}{2}$ and see what you get for $h$

anonymous · Answer

Wait, sorry I'm still a bit confused. @satellite73

anonymous · Answer

satellites way is the only way unless you know calculus

anonymous · Answer

the max height of the ball is at the vertex of the parabola

anonymous · Answer

Oh, that's probably why then. I haven't learned calculus yet.

anonymous · Answer

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