# Answer to Question #71812 in Mechanics | Relativity for Bianca

Question #71812

A ball is thrown directly downward with an initial speed of 8 m/sec, from a height of 30 m. After what time interval does it strike the ground?

Expert's answer

x = v0t + (1/2) at^2

30m = 8. 0t + (1/2) 9.8t^2

0 = 4.9t2 + 8.0t – 30

We find the discriminant of the square equation:

D = b2 - 4ac = 82 - 4 · 4.9 · (-30) = 64 + 588 = 652

Since the discriminant is greater than zero, then the square equation has two real roots:

t1 = -8 - √652 / 2·(4.9) = -4049 - 1049√163 ≈ -3.4218663948578985

Since the negative number makes no sense,

t2 = -8 + √652 / 2·(4.9 = -4049 + 1049√163 ≈ 1.7892133336334088 ≈ 1.79 second

30m = 8. 0t + (1/2) 9.8t^2

0 = 4.9t2 + 8.0t – 30

We find the discriminant of the square equation:

D = b2 - 4ac = 82 - 4 · 4.9 · (-30) = 64 + 588 = 652

Since the discriminant is greater than zero, then the square equation has two real roots:

t1 = -8 - √652 / 2·(4.9) = -4049 - 1049√163 ≈ -3.4218663948578985

Since the negative number makes no sense,

t2 = -8 + √652 / 2·(4.9 = -4049 + 1049√163 ≈ 1.7892133336334088 ≈ 1.79 second

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