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).
1 answer