A ball is thrown from a height of
130
feet with an initial downward velocity of
6/fts
. The ball's height
h
(in feet) after
t
seconds is given by the following.
=h−130−6t16t2
How long after the ball is thrown does it hit the ground?
To find out how long it takes for the ball to hit the ground, we need to find the value of t when the height of the ball, h, is equal to 0.
So, we need to solve the equation:
0 = h - 130 - (6t/ft) + (16t^2)
Simplifying the equation, we get:
16t^2 - (6t/ft) - 130 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, the equation is not easily factored, so let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = 16, b = -6/ft, and c = -130.
Substituting the values into the formula, we have:
t = (-(-6/ft) ± √((-6/ft)^2 - 4 * 16 * -130)) / (2 * 16)
Simplifying further:
t = (6/ft ± √((36/ft^2) + 8320)) / 32
t = (6/ft ± √((36 + 8320ft^2)/ft^2)) / 32
To simplify the square root term, let's first calculate the discriminant, which is the value inside the square root:
discriminant = (36 + 8320ft^2)/ft^2
Now, we can substitute the discriminant into the equation:
t = (6/ft ± √(discriminant)) / 32
To find out how long after the ball is thrown it hits the ground, we need to solve for the positive value of t when h = 0. We can ignore the negative value because time cannot be negative in this context.
So, in the equation t = (6/ft ± √(discriminant)) / 32, we only consider the positive value.
Therefore, the time it takes for the ball to hit the ground is given by:
t = (6/ft + √(discriminant)) / 32
Please note that to determine the exact value of t and work with appropriate units, we need to know the exact value of the discriminant by substituting the given units for h, t, and ft into the equation.
To find how long after the ball is thrown does it hit the ground, we need to determine the time when the height, given by the function h(t), becomes zero.
Given that the height function is given by:
h(t) = -16t^2 - 6t + 130
To find the time when the ball hits the ground, we set h(t) to zero and solve for t:
0 = -16t^2 - 6t + 130
This equation is a quadratic equation, so we can use the quadratic formula to solve it:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = -16, b = -6, and c = 130
Plugging in these values into the quadratic formula, we get:
t = (-(-6) ± √((-6)^2 - 4(-16)(130))) / (2(-16))
t = (6 ± √(36 + 8320)) / (-32)
t = (6 ± √(8356)) / (-32)
t = (6 ± 91.39) / (-32)
Solving further:
t = (6 + 91.39) / (-32) or t = (6 - 91.39) / (-32)
t = 97.39 / (-32) or t = -85.39 / (-32)
t ≈ -3.04 or t ≈ 2.67
Since time cannot be negative for this problem, we discard the negative solution and consider the positive solution. Therefore, the ball hits the ground approximately 2.67 seconds after it is thrown.
Your equation should have been
h = -16t^2 - 6t + 130
When it hits the ground, h = 0
so,
16t^2 + 6t - 130 = 0
I would use the quadratic equation formula to solve for t
(to be able to handle this topic, you MUST know that formula)
make sure to reject any negative value of t.