SOLUTION: A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function . h(t)=100t-16t^2 What is the maximum height that the ball will reach?

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Question 697309: A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function .
h(t)=100t-16t^2
What is the maximum height that the ball will reach? Height= ? feet
Do not round your answer.
Found 2 solutions by josmiceli, nerdybill:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Plot this function with +h%28t%29+ on the vertical axis
and +t+ on the horizontal axis.
This is a parabola that has a maximum. The maximum is
half-way between the 2 roots. One of the roots is at
(0,0) at the instant the ball was thrown upward.
Find the other root by setting +h%28t%29+=+0+ , and
find the time when the ball comes back to the ground.
+100t+-+16t%5E2+=+0+
+50t+-+8t%5E2+=+0+
+t%2A%28+50+-+8t+%29+=+0+
+50+-+8t+=+0+
+8t+=+50+
+t+=+25%2F4+
The time half-way between +0+ and +25%2F4+
is +25%2F8+. Now find the height at +t+=+25%2F8+
+h%28t%29+=+100t+-+16t%5E2+
+h%2825%2F8%29+=+100%2A%2825%2F8%29+-+16%2A%2825%2F8%29%5E2+
+h%2825%2F8%29+=+12.5%2A25+-+16%2A%28625%2F64%29+
+h%2825%2F8%29+=+312.5+-+625%2F4+
+h%2825%2F8%29+=+312.5+-+156.25+
+h%2825%2F8%29+=+156.25+
The maximum height is 156.25 ft
Here's the plot:
+graph%28+400%2C+400%2C+-2%2C+8%2C+-20%2C+200%2C+-16x%5E2+%2B+100x+%29+



Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Max height is at the vertex:
the time (t) of the vertex is:
t = -b/(2a)
t = -100/(2(-32))
t = -100/(-64)
t = 100/64
t = 25/16
t = 1.5625 seconds
.
Max height is:
h(t)=100t-16t^2
h(1.5625)=100(1.5625)-16(1.5625)^2
h(1.5625)=156.25-39.0625
h(1.5625)=117.1875 feet