SOLUTION: height of an object thrown from 60 feet up, with an initial speed of 48 feet/second is given by h(t)=-16t^2+48t+60. When will the object reach its maximum height? What will the max

Question 1052691: height of an object thrown from 60 feet up, with an initial speed of 48 feet/second is given by h(t)=-16t^2+48t+60. When will the object reach its maximum height? What will the maximum height be?
Found 2 solutions by htmentor, josmiceli:
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
The path of the object is a downward facing parabola. The object will reach its maximum height when dh(t)/dt = 0.
dh/dt = -32t + 48 = 0
This gives t = 1.5 seconds
The maximum height will be the value of the function at t = 1.5:
h(1.5) = 96 feet
The path of the object is given below:

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
The t-value of the vertex ( maximum height ) is
goiven by the formula:
, where:

Plug this result back into equation

The maximum height is 96 ft
----------------------------
Here's the plot:

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