THe function h(t)=-16t^2+v0t+h0 describes the height in feet above the ground h(t) of an object thrown vertically from a height of h0 feet, with an initial velocity of v0 feet per second, if there is no air friction and t is the time in seconds sine the object was thrown. A ball is thrown upward from a 100-foot tower at a velocity of 60 feet per second. How many seconds will it take for the ball to reach the ground?

The real question is who can throw that hard?

Well, I must say, that ball must really have had some lofty aspirations if it wants to reach all the way to the ground from the 100-foot tower! Now, let's solve this conundrum.

We know that when the ball reaches the ground, its height above the ground will be zero. So, we need to find the time at which the height function, h(t), equals zero.

In this case, h(t) = -16t^2 + v0t + h0, where h0 is the initial height of the tower, v0 is the initial velocity, and t is the time.

Let's plug in the values: h(t) = -16t^2 + 60t + 100.

To find the time it takes for the ball to reach the ground, we need to find the value of t when h(t) equals zero.

So, we have -16t^2 + 60t + 100 = 0.

Now, to solve this equation, we can either use the quadratic formula or factor it. But since we're talking about a flying ball here, let's give it a little extra air time and factor it.

After some fancy factorization tricks (I bet you wish you could juggle like that!), we get (4t + 10)(-4t + 10) = 0.

Simplifying further, we have t = -10/4 and t = 10/4.

Now, since we're dealing with time, we can't have negative values. So, we discard t = -10/4.

Finally, we have t = 10/4, which simplifies to t = 2.5 (seconds).

Therefore, the ball will take 2.5 seconds to make its way back down to terra firma.

I hope that answers your question without bouncing off course too much!

To find the time it takes for the ball to reach the ground, we need to determine when the height, h(t), is equal to zero.

Given:
h(t) = -16t^2 + v0t + h0
v0 = 60 (initial velocity)
h0 = 100 (initial height)

We want to find t when h(t) = 0.

Plugging in the given values into the equation:
0 = -16t^2 + 60t + 100

This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:

The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

Applying this formula to our equation, where a = -16, b = 60, and c = 100:

t = (-60 ± √(60^2 - 4*(-16)*100)) / 2*(-16)

Simplifying:

t = (-60 ± √(3600 + 6400)) / -32
t = (-60 ± √10000) / -32
t = (-60 ± 100) / -32

We have two possibilities:

1. t = (-60 + 100) / -32 = 40 / -32 = -1.25
2. t = (-60 - 100) / -32 = -160 / -32 = 5

In our case, we can discard the negative solution since time cannot be negative. Therefore, the ball will take 5 seconds to reach the ground.

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The answer is 5 seconds.

-16t^2+60t+100+0