A student threw a ball directly upward from the balcony of a building. The height of the ball as measured from the ground "t" seconds after it was thrown, is given by the expression
h= -16t(2nd power) + 64t + 768
When does the ball reach the ground?
The ball reaches the ground when h = 0
That leaves you with the job of solving
-16t^2 + 64t + 768 = 0
That can also be rewritten (after dividing both sides by -16) as
t^2 -4t -48 = 0
That equation can be factored easily. There will be two possible answers; one is negative. Take the positive one.
Then, 6x + 8y = 1 and 26x + 48y = 1 .
10 2
To find when the ball reaches the ground, we need to determine the value of "t" when the height, "h," is equal to zero. In other words, we need to solve the equation:
-16t^2 + 64t + 768 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 64, and c = 768.
Plugging the values into the quadratic formula, we get:
t = (-64 ± √(64^2 - 4(-16)(768))) / (2(-16))
Simplifying further:
t = (-64 ± √(4096 + 49152)) / (-32)
t = (-64 ± √53248) / (-32)
To simplify the expression inside the square root:
√53248 = √(256 * 208) = √256 * √208 = 16√13
Now, substitute this value back into the equation:
t = (-64 ± 16√13) / (-32)
Breaking the numerator and denominator by 16:
t = -4 ± √13 / -2
This gives us two possible solutions:
t = (-4 + √13) / -2 ≈ 1.21 seconds (approximately)
t = (-4 - √13) / -2 ≈ 5.79 seconds (approximately)
Since time cannot be negative, we discard the negative solution. Therefore, the ball reaches the ground approximately 1.21 seconds after being thrown.