SOLUTION: If an object is propelled upward from a 288 foot high tower at an initial velocity of 48 feet per second, then its height h(in feet) after t seconds is given by the equation h=-16t
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-> SOLUTION: If an object is propelled upward from a 288 foot high tower at an initial velocity of 48 feet per second, then its height h(in feet) after t seconds is given by the equation h=-16t
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Question 172759: If an object is propelled upward from a 288 foot high tower at an initial velocity of 48 feet per second, then its height h(in feet) after t seconds is given by the equation h=-16t^2+48t+288. What is the maximum height reached by the object? Answer by solver91311(24713) (Show Source):
1. Algebra solution. The vertex of a parabola described by has an x-coordinate at and a y-coordinate at If then you have a concave up parabola and the vertex represents a minimum value of the function. If , then it is concave down and the vertex represents a maximum. The latter is the situation in the given problem.
That means the t-coordinate of the vertex is at
Evaluating to determine the h-coordinate of the vertex which is also the maximum height:
Maximum height: 324 feet
2. Calculus method.
A local extrema for a continuous differentiable function is located where the value of the first derivitive is equal to zero. If the function is twice differentiable, the character of the extrema (maximum or minimum) can be determined by examining the sign of the second derivative evaluated at the extreme point.
(Look familiar?)
Evaluation of is the same as in the Algebra method above, so the value for maximum height is the same, namely 324 feet.
which is negative for all values of t in the domain of h(t), therefore the extreme point is a maximum.
