A flying cannonball’s height is described by formula y=−16t^2+300t. Find the highest point of its trajectory. In how many seconds after the shot will cannonball be at the highest point?
Question
Answer:
Highest point = 1406.25
Number of seconds = 9.375
We've been given the quadratic equation y = -16t^2 + 300t which describes a parabola. Since a parabola is a symmetric curve, the highest value will have a t value midway between its roots. So using the quadratic formula with A = -16, B = 300, C = 0. We get the roots of t = 0, and t = 18.75. The midpoint will be (0 + 18.75)/2 = 9.375
So let's calculate the height at t = 9.375.
y = -16t^2 + 300t
y = -16(9.375)^2 + 300(9.375)
y = -16(87.890625) + 300(9.375)
y = -1406.25 + 2812.5
y = 1406.25
So the highest point will be 1406.25 after 9.375 seconds.
Let's verify that. I'll use the value of (9.375 + e) for the time and substitute that into the height equation and see what I get.'
y = -16t^2 + 300t
y = -16(9.375 + e)^2 + 300(9.375 + e)
y = -16(87.890625 + 18.75e + e^2) + 300(9.375 + e)
y = -1406.25 - 300e - 16e^2 + 2812.5 + 300e
y = 1406.25 - 16e^2
Notice that the only term with e is -16e^2. Any non-zero value for e will cause that term to be negative and reduce the total value of the equation. Therefore any time value other than 9.375 will result in a lower height of the cannon ball. So 9.375 is the correct time and 1406.25 is the correct height.
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