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

Algebra ->  Quadratic Equations and Parabolas -> 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      Log On


   

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) About Me  (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:
+graph%28+400%2C+260%2C+-1%2C+4%2C+-10%2C+100%2C+-16x%5E2%2B48x%2B60+%29

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The t-value of the vertex ( maximum height ) is
goiven by the formula:
+t%5Bmax%5D+=+-b%2F%282a%29+, where:
+a+=+-16+
+b+=+48+
+t%5Bmax%5D+=+-48%2F%282%2A%28-16%29%29+
+t%5Bmax%5D+=+24%2F16+
+t%5Bmax%5D+=+3%2F2+
Plug this result back into equation
+h%283%2F2%29+=+-16%2A%283%2F2%29%5E2+%2B+48%2A%283%2F2%29+%2B+60+
+h%283%2F2%29+=+-16%2A%289%2F4%29+%2B+72+%2B+60+
+h%283%2F2%29+=+-36+%2B+132+
+h%283%2F2%29+=+96+
The maximum height is 96 ft
----------------------------
Here's the plot:
+graph%28+400%2C+400%2C+-1%2C+5%2C+-15%2C+125%2C+-16x%5E2+%2B+48x+%2B+60+%29+