SOLUTION: Hello, I need help with this problem:
An object is tossed into the air with the upward velocity of 22 feet per second from the top of a 10-foot wall. Its flight is governed by h
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Question 455237: Hello, I need help with this problem:
An object is tossed into the air with the upward velocity of 22 feet per second from the top of a 10-foot wall. Its flight is governed by h(t)=16t^2-22t-10=0. What is the time for the object to reach the highest point, and what is the altitude of this highest point?
Thank you.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
An object is tossed into the air with the upward velocity of 22 feet per second from the top of a 10-foot wall. Its flight is governed by h(t)=16t^2-22t-10=0. What is the time for the object to reach the highest point, and what is the altitude of this highest point?
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h(t)=16t^2-22t-10 = 0
If it equals 0, there's not much solving to do.
h(t)=16t^2-22t-10
That's a parabola that opens upward. h increases without limit with increasing time. There would be no highest point.
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Use h(t)= -16t^2 + 22t + 10, -1 times the orginal.
h(t)= -16t^2 + 22t + 10
The vertex is on the line of symmetry.
The LOS is t = -b/2a
t = -22/-32 = 11/16 seconds, the time at apogee, the highest point.
h(11/16) = -16*(11/16)^2 + 22*(11/16) + 10
= -121/16 + 121/8 + 10
= 281/16 feet
= 17.5625 ft
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