Jesse throws a basketball such that its height is modelled by the function h(t) = −4.9(t – 1.5)2 + 12.5, where h is the height in metres and t is time in seconds.
What is the maximum height of the ball?
How many seconds is the ball in the air?
The maximum height of the ball can be found by finding the vertex of the quadratic function. The vertex is at (1.5, 12.5). Therefore, the maximum height of the ball is 12.5 meters.
To find how many seconds the ball is in the air, we need to find the value of t when the height is 0 (since the ball is on the ground at that point). Setting h(t) = 0 and solving for t, we get:
0 = −4.9(t – 1.5)2 + 12.5
Rearranging the equation, we have:
4.9(t – 1.5)2 = 12.5
Dividing both sides by 4.9, we get:
(t – 1.5)2 = 12.5/4.9
Taking the square root of both sides, we have:
t – 1.5 = ±√(12.5/4.9)
Simplifying further, we have:
t = 1.5 ± √(12.5/4.9)
Evaluating this equation, we find two solutions:
t ≈ 1.5 + √(2.551) ≈ 1.5 + 1.6 ≈ 3.1 seconds
t ≈ 1.5 - √(2.551) ≈ 1.5 - 1.6 ≈ -0.1 seconds
Since time cannot be negative, we can ignore the second solution. Therefore, the ball is in the air for approximately 3.1 seconds.