To find the time at which the height of the soccer ball is -65, we need to set the function h(t) equal to -65 and solve for t:
16t² + 98t = -65
Rearrange the equation:
16t² + 98t + 65 = 0
Now we can use the quadratic formula to solve for t:
t = (-b ± √(b² - 4ac)) / (2a)
In this case, a = 16, b = 98, and c = 65.
t = (-98 ± √(98² - 4(16)(65))) / (2(16))
t = (-98 ± √(9604 - 4160)) / 32
t = (-98 ± √5444) / 32
Since we are interested in the time, we ignore the negative value:
t = (-98 + √5444) / 32
t ≈ -0.21 (rounded to two decimal places)
So, the time at which the height of the soccer ball is -65 is approximately -0.21 seconds.
The function h (t) = 16t² + 98t graphed below shows the height, in feet, of a kicked soccer ball as a function of time, in seconds.
-701
-65
60
55
50
45
40
35
30
25
20
15
10
5
1 answer
1 answer