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Question

A gun fires a bullet almost straight up from the edge of a 140-foot cliff. If the bullet leaves the gun with a speed of 556 feet per second, its height at time t is given by h(t)=−16t2+556t+140, measured from the ground below the cliff.

When will the bullet land on the ground below the cliff? (Hint: What is its height when it lands? Remember that we are measuring from the ground below, not from the cliff.) 
The bullet will land after  seconds.

electron1 · Answer

Let me show you the equation that is used in physics to solve this type of problem.

hf = hi + vi * t – ½ * g * t^2

hf is the object’s final height. In this problem, hf = 0 ft. hi is the initial height of the object. In this problem, hi is 140 ft. vi is the object’s initial upward velocity. In this problem, vi is 556 ft/s. g is the acceleration toward the ground. Let’s set ½ * g equal to 16 and solve for g.

½ * g = 16, g = 32 ft/s^2
This means the bullet’s velocity is decreasing at the rate of 32 ft/s.

Let’s put all these numbers into the physics equation and then solve for t.

0 = 140 + 556 * t – ½ * 32 * t^2
16 * t^2 – 556 * t – 140 = 0
Solve for t
I use the following website to solve quadratic equations. The time is 35 seconds.
http://www.math.com/students/calculators/source/quadratic.htm

I hope my explanation and the physics equation helps you to understand how to solve his type of problem.

nosi · Answer

The bullet will be at 0 feet when it lands, since we're measuring from the lower ground. So what you need to do is set the equation to zero and solve for t.

−16t^2+556t+140 = 0
t = [-556 +- sqrt(556^2 - 4(-16)(140))]/-32
t = [-556 +- sqrt(318096)]/-32
t = [-556 +- 564]/-32
t = 35, or t = -1/4

We can throw out the negative answer, and t = 35 seconds
You can also find this by punching y=−16t^2+556t+140 in your calculator; look at the table, and see that when y = 0, x = 35. I checked.
Hope this helped!

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Help with this math problem?

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