A ball is thrown upward with an initial velocity of 14 meters per second from a cliff that is high. The height of the ball is given by the quadratic equation where h is in meters and t is the time in seconds since the ball was thrown. Find the time that the ball will be 30 meters from the ground. Round your answer to the nearest tenth of a second.
PS: I know the answer, because it was a test question, but I do not know how to come up with it. Please show me steps, I am going crazy here. Thanks for your time:)
What is the ht. 0f the cliff?
To find the time that the ball will be 30 meters from the ground, we need to solve the given quadratic equation for h = 30.
The quadratic equation representing the height of the ball is given by:
h = -4.9t^2 + 14t
Substituting h = 30 into the equation, we get:
30 = -4.9t^2 + 14t
To solve this quadratic equation, we can rearrange it to the standard form:
4.9t^2 - 14t + 30 = 0
We can solve this equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation:
a = 4.9, b = -14, and c = 30.
Let's calculate the values and solve the equation step-by-step:
1. Calculate the discriminant:
D = b^2 - 4ac
D = (-14)^2 - 4 * 4.9 * 30
2. Calculate the square root of the discriminant:
√D = √((-14)^2 - 4 * 4.9 * 30)
3. Calculate the two possible solutions of t:
t1 = (-(-14) + √D) / (2 * 4.9)
t2 = (-(-14) - √D) / (2 * 4.9)
4. Calculate the approximate values of t1 and t2 rounded to the nearest tenth of a second.
By following these steps, you should be able to find the time at which the ball will be 30 meters from the ground.
To find the time that the ball will be 30 meters from the ground, we can set the height equation equal to 30 and solve for time.
The height equation is given as:
h = -16t^2 + 14t + h₀
Where:
h = height of the ball at time t
t = time in seconds since the ball was thrown
h₀ = initial height of the ball
In this case, the initial height of the ball from the cliff is not given, but we can assume it to be zero since it is thrown vertically upwards.
So, the height equation becomes:
h = -16t^2 + 14t
We want to find the time when the height is 30 meters, so we set h equal to 30 and solve for t:
30 = -16t^2 + 14t
Rearranging the equation, we have:
16t^2 - 14t + 30 = 0
This is a quadratic equation in standard form (ax^2 + bx + c = 0), where:
a = 16
b = -14
c = 30
To solve the quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
t = (-(-14) ± √((-14)^2 - 4 * 16 * 30)) / (2 * 16)
t = (14 ± √(196 - 1920)) / 32
t = (14 ± √(-1724)) / 32
Since √(-1724) is imaginary (no real square root of a negative number), we can conclude that there are no real solutions to this equation.
Therefore, there is no time at which the ball will be exactly 30 meters from the ground.