5.) A ball is thrown into the air with an initial velocity of 96 feet per second. Given the formula below wheret,
represents time in seconds and s(t), the height of the ball, when will the ball hit the ground?
s(t) = -16t? + 960

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The time required to hit the ground is required.

The time taken by the ball to reach the ground is 2.45 seconds.

The equation is

At the ground

so,

The time taken by the ball to reach the ground is 2.45 seconds.

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The ball will hit the ground 6 seconds after it has ben thrown, at t = 6s.

Step-by-step explanation:

The equation for the height of the ball should be:

s(t) = (-16ft/s^2)*t^2 + 96ft/s*t

When s(t) = 0ft, means that the ball is in the ground

This represents the height of the ball as a function of t, the time in seconds.

We can see that at t = 0s:

s(0s) = (-16ft/s^2)*(0s)^2 + 96ft/s*(0s) = 0ft

Then at the time 0 seconds, the ball is in the ground, so we must look at the other root of the equation

(-16ft/s^2)*t^2 + 96ft/s*t = 0ft

To find it, we can use the Bhaskara equation, in this case is:

Then we have the two solutions:

t = (-96ft/s + 96ft/s)/(-32 ft/s^2) = 0s  (the one that we already found)

And the other one is:

t = (-96ft/s - 96ft/s)/(-32ft/s^2) = 6s

Then:

s(6s) = 0ft

This means that the ball will hit the ground at t = 6 seconds.

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