A ball is thrown directly downward, with an initial speed of 6.10 m/s, from a height of 31.0 m. After what interval does the ball strike the ground?

d = Vi*t + 0.5gt^2 = 31m,

6.1t + 0.5*9.8t^2 = 31,
4.9t^2 + 6.1t - 31 = 0,
Solve for t using Quad. Formula and get:

t = 1.97s.

To find the time it takes for the ball to strike the ground, we can use the kinematic equation for vertical motion:

h = ut + (1/2) gt^2

Where:
h = initial height = 31.0 m (negative since it is measured downward)
u = initial velocity = -6.10 m/s (negative since it is directed downward)
g = acceleration due to gravity = -9.8 m/s^2 (negative since it is directed downward)
t = time of flight (what we need to find)

To solve for t, we rearrange the equation:

31.0 = (-6.10)t + (1/2)(-9.8)t^2

Simplifying the equation, we get:

-4.9t^2 - 6.10t + 31.0 = 0

This is a quadratic equation, which can be solved using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -4.9, b = -6.10, and c = 31.0. Substituting these values into the quadratic formula, we have:

t = (-(-6.10) ± √((-6.10)^2 - 4(-4.9)(31.0))) / (2(-4.9))

Simplifying further, we get:

t = (6.10 ± √(37.21 + 607.6)) / (-9.8)

Calculating inside the square root:
√(37.21 + 607.6) = √644.81 = 25.4

Substituting the values back into the equation, we have:

t = (6.10 ± 25.4) / (-9.8)

Now we can solve for the two possible values of t:

Case 1: t = (6.10 - 25.4) / (-9.8) = -19.3 / (-9.8) ≈ 1.97 s
Case 2: t = (6.10 + 25.4) / (-9.8) = 31.5 / (-9.8) ≈ -3.22 s

Since time cannot be negative in this context, we discard the negative value and consider the positive value. Therefore, the ball strikes the ground approximately after 1.97 seconds.