A ball is thrown into the air with the in initial upward velocity of 45 ft/s. It's height (h) in feet after t seconds is given by the function h=- -16t^2+45t+6. After about how many seconds will the ball hit the ground?
my answer is 3 seconds
To find out how many seconds it will take for the ball to hit the ground, we need to determine the value of t when h = 0. In other words, we need to find the time at which the height of the ball is zero.
The given function that represents the height of the ball is h = -16t^2 + 45t + 6. We can set this equal to zero:
-16t^2 + 45t + 6 = 0
To solve this quadratic equation, we can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = -16, b = 45, and c = 6. Plugging these values into the quadratic formula, we get:
t = (-45 ± √(45^2 - 4*(-16)*6)) / (2*(-16))
Simplifying further:
t = (-45 ± √(2025 + 384)) / (-32)
t = (-45 ± √(2409)) / (-32)
Now, we can calculate the values of t using both the positive and negative values in the formula:
t1 = (-45 + √(2409)) / (-32)
t2 = (-45 - √(2409)) / (-32)
Let's calculate these values:
t1 ≈ 4.16 seconds
t2 ≈ -0.34 seconds
Since time cannot be negative in this context, we disregard the negative solution and take the positive value as the approximate time it will take for the ball to hit the ground.
Therefore, after approximately 4.16 seconds, the ball will hit the ground.