A ball is thrown directly downward with an initial speed of 8.50 m/s, from a height of 29.7 m. After what time interval does it strike the ground?

Using the following formula for linear-motion, the missing variable can be solved:

s = Vi * t + 1/2 (a * t^2)

Where: s = displacement = 29.7 m

Vi = initial velocity = 8.5 m/s

a = acceleration = 9.8

t = time = ?

Substituting:

29.7 = 8.5t + 1/2 (9.8*t^2)

29.7 = 8.5t + 4.9t^2

Dividing both sides by 4.9:

6.06 = 1.73t + t^2

t^2 + 1.73t - 6.06 = 0

(t - 1.74) (t + 3.48) = 0

t = 1.74s

From the above values, the correct answer is 1.74 seconds.