I am stuck form this point i need help.
the problem states:
A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s. Use this information for exercises 35 to 38.
Science and medicine. When will the ball reach a height of 80 m?
h=-5t^2+20t+100
therefore this is what i did:
80=-5t^2+20t+100
0=-5t^2+20t+100-80
0=-5t^2+20t+20
0=-1(-5t^2+20t+20)
0=5t^2-20t-20
Now from this point can someone show me how to finish this.
so i can know how these types of problems have to look
first of all divide all terms by 5 to reduce it
Since you now have a quadratic equation which does not factor, I assume you know about the quadratic equation
I got t = 2+√8 or 2-√8,
I leave it up to you to decipher the validity of those answers.
remember that when t-0, h = 100
A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s. Use this information for exercises 35 to 38.
When will the ball reach a height of 80 m?
1- The ball will reach a height of h m in "t1" seconds.
2-From Vf = Vo - gt1, 0 = 20 - 9.8t1 making t1 = 2.0408 seconds.
3-The height traveled in this time period is given by h = Vot - 4.9t1^2 = 20(2.0408) - 4.9(2.0408)^2 = 20.408 m.
3- The ball now falls from a height of 120.408m to 80m in t2 sec derived from 140.408 = 0 + 4.9t2^2 or t2 = sqrt(120.408/4.9) = 5.35 sec.
4- The total time taken to reach its maximum height and falling back down to 80m is therefore 2.0408 + 5.35 = 7.3908 sec.
C97,3).
C97,3).
To solve the problem, you have correctly set up the equation:
80 = -5t^2 + 20t + 100
To finish solving the equation, you can follow these steps:
1. Simplify the equation:
0 = 5t^2 - 20t - 20
2. Divide all terms by 5 to reduce it:
0 = t^2 - 4t - 4
3. Since this equation does not factor easily, you can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -4, and c = -4. Plugging these values into the formula, you get:
t = (-(-4) ± √((-4)^2 - 4(1)(-4))) / (2(1))
t = (4 ± √(16 + 16)) / 2
t = (4 ± √32) / 2
t = (4 ± 4√2) / 2
t = 2 ± 2√2
Simplify:
t1 = 2 + 2√2
t2 = 2 - 2√2
These are the two possible solutions for t. You can check their validity by plugging them back into the original equation.
Remember that when t = 0, h = 100, so the positive value of t, t1, represents the first time the ball reaches the desired height.
Therefore, the ball will reach a height of 80 m after approximately 2 + 2√2 seconds (t1) from the time it was thrown.