During intermission, at a hockey game, small foam hockey pucks are launched from a height. How long is a puck in the air if a student in the stands catches it on its way down 12m above ice lvl? The model for the vertical motion of the puck can be approximated by the relation h=-5t^2+15t+2, where t is the time in seconds and h is the height in metres.
well, just solve for t when
-5t^2 + 12t + 2 = 12
But I suspect a typo, since the maximum height reached is 9.2m
??
To find out how long the puck is in the air before it reaches a height of 12m above the ice level, we can set up an equation using the given model for the vertical motion of the puck.
The equation is h = -5t^2 + 15t + 2, where h is the height in meters and t is the time in seconds.
We need to find the value of t when the height h is 12m. So, we can substitute h = 12 into the equation:
12 = -5t^2 + 15t + 2
Now, we have a quadratic equation. To solve it, we can rearrange it to bring it to standard quadratic form:
-5t^2 + 15t + 2 - 12 = 0
-5t^2 + 15t - 10 = 0
Now, we can solve this equation by factoring, completing the square, or using the quadratic formula.
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
In this equation, a = -5, b = 15, and c = -10.
Plugging these values into the quadratic formula:
t = (-15 ± √(15^2 - 4(-5)(-10))) / (2(-5))
Simplifying:
t = (-15 ± √(225 - 200)) / (-10)
t = (-15 ± √25) / (-10)
Now, we have two possible solutions:
t1 = (-15 + 5) / (-10) = -10 / -10 = 1
t2 = (-15 - 5) / (-10) = -20 / -10 = 2
Both t = 1 and t = 2 give us valid solutions for the time. However, since we are looking for the time during the puck's descent, we would consider the positive value, t = 2.
Therefore, the puck is in the air for 2 seconds before it reaches a height of 12m above the ice level.