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.
How many seconds is the ball in the air?
To determine how many seconds the ball is in the air, we need to find the value of t when the height of the ball is 0.
Setting h(t) = 0, we can solve for t:
0 = -4.9(t - 1.5)^2 + 12.5
Rearranging this equation, we get:
4.9(t - 1.5)^2 = 12.5
Dividing both sides of the equation by 4.9, we have:
(t - 1.5)^2 = 12.5/4.9
(t - 1.5)^2 ≈ 2.551
Taking the square root of both sides, we get:
t - 1.5 ≈ ±√2.551
t ≈ 1.5 ± √2.551
Since time cannot be negative in this context, we can ignore the negative square root:
t ≈ 1.5 + √2.551
t ≈ 1.5 + 1.595
t ≈ 3.095
Therefore, the ball is in the air for approximately 3.095 seconds.