Question 863027: A baseball is hit from a height of four feet off the ground. Its height 'h' (in feet) depends on the time 't' (in seconds) that it has been in the air. It can be modeled by the equation: h(t)=-16t^2+30t+4
a.) How long does it take until the ball hits the ground?
b.) When does it reach its highest height?
c.) What is its highest height?
d.) sketch the graph of the h(t). Be sure to label your axis, and include all intercepts and the vertex.
My work so far:
a.)0=-16t^2+30t+4
2(-8t^2+15t+2)
2(-8t-1)(t-2)
t= -1/8 and 2 seconds, respectively- This means the ball hits the ground after two seconds because -1/8 of a second cannot happen.
b.)h(0)=-16(0)^2+30(0)+4
0+0+4
4 seconds is how long it takes for the ball to reach its highest height but that cannot happen if it hit the ground after two seconds.
c.) cannot solve until I figure out where I went wrong in the beginning.
d.) I do not really need help with this one its just part of the problem so I included it but this is the only part I actually know how to do.
Thank you!
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! Your solution for part a is correct.
Your work for part b is correct but not for that question.
You calculated the height at time 0
You substituted 0 for t which is time and found h
You want to find the vertex ie the highest point.
There are two ways to do that.
1) Find t or x of the vertex by x=-b/2a
-30/-32=15/16=t
or
2) by finding the first derivative
d/dt(-16t^2+30t+4) = -32t+30
32t=30
t=30/32=15/16
so after 15/16 seconds the ball is at its highest point.
c) Plug 15/16 to find h(t)
vertex | (15/16, 289/16)=(0.9375, 18.0625)
The ball's highest point is 18 .0625 feet after .9375 seconds
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