a ball is thrown into the air at a speed of 96 feet/sec. The equation that expresses the height of the ball as a function of time is: h(t) = -16t2 + 96t + 5
How long does the ball stay in the air after it is thrown? (Round answer to the hundreths of a second)
To determine how long the ball stays in the air after it is thrown, we need to find the time when the height of the ball is zero. In other words, we need to find the value of t when h(t) = 0.
Given the equation that expresses the height of the ball as a function of time: h(t) = -16t^2 + 96t + 5
Setting h(t) = 0, we get:
0 = -16t^2 + 96t + 5
Now we can solve this quadratic equation for t by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Given: a = -16, b = 96, c = 5
t = (-(96) ± √((96)^2 - 4(-16)(5))) / (2(-16))
Simplifying further:
t = (-96 ± √(9216 + 320)) / (-32)
t = (-96 ± √9536) / (-32)
t = (-96 ± 97.65) / (-32)
We have two possible solutions for t:
t₁ = (-96 + 97.65) / (-32) ≈ -0.023 seconds
t₂ = (-96 - 97.65) / (-32) ≈ 6.024 seconds
Since the time cannot be negative in this scenario, we discard the negative solution.
Therefore, the ball stays in the air for approximately 6.024 seconds (rounded to the hundredths of a second).