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Duane throws a tennis ball in the air. After 1 second, the height of the ball is 53 ft.
After 2 seconds, the ball reaches a maximum height of 69 feet.
After 3 seconds the height of the ball is 53 ft.
Part A. Write a quadratic function to represent the height of the ball, ℎ, at any point in time, 𝑡.
Part B. How long will the tennis ball stay in the air? Round your answer to the nearest
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We are given that at t= 2 seconds the height is maximum 69 feet.
It means that the quadratic function in vertex form has the formula
h(t) = a*(t-2)^2 + 69.
In this formula, we have only one unknown parameter "a".
To find it, we use the condition h(1) = 53
53 = a*(1-2)^2 + 69.
It gives
a*(-1)^2 = 53 - 69
a = -16.
So, the quadratic function of the height is
h(t) = -16*(t-2)^2 + 69. <<<---=== it is the ANSWER to question (A).
At t= 0 (starting moment), the starting height is
h(0) = -16*(0-2)^2 + 69 = -16*4 + 69 = 5 ft.
The ball will hit the ground when h(t) = 0
-16*(t-2)^2 + 69 = 0
-16*(t-2)^2 = - 69
16(t-2)^2 = 69
(t-2)^2 = 69/16
t - 2 = = 2.0767 seconds (rounded)
t = 2 + 2.0767 = 4.0767 seconds (rounded).
The ball is in the air 4.0767 seconds.
It is the ANSWER to question (B).
Solved.