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
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
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!