A helicopter pilot drops a package when the helicopter is 200 ft. above the ground, rising at 20 ft/sec.
How long will it take for the package to hit the ground? What is the speed of the package at impact?
I know you set s(t) equal to zero, so -16(t^2)+20t+200=0, but factoring it out I get really obscure values for t. I'm wondering if there's another way to do it.
-16t^2+20t+200=0
divide by -4
4t^2 - 5t - 50 = 0
This does not factor, so use the formula
t = (5 ± √825)/8
= 4.2 or a negative
but since t > 0
t = 4.2 seconds
To find the time it takes for the package to hit the ground, we can use the quadratic equation. However, factoring the equation might not always yield nice and simple solutions.
Alternatively, we can use the concept of projectile motion to solve this problem. When an object is in free fall, the equation s(t) = -16t^2 + v0 * t + s0 can be used, where s(t) is the height of the object at time t, v0 is the initial velocity, and s0 is the initial height.
In this case, the initial height is 200 ft, and the initial velocity can be found by realizing that the package is rising at 20 ft/sec. Therefore, the initial velocity is -20 ft/sec.
Plugging the values into the equation, we have:
s(t) = -16t^2 - 20t + 200
To find when the package hits the ground, we want to find the time, t, where s(t) is equal to zero. So we set s(t) = 0 and solve for t:
-16t^2 - 20t + 200 = 0
Now, we can solve this quadratic equation either by factoring or by using the quadratic formula.
If you prefer factoring, keep in mind that the quadratic equation may not always factor nicely. In this case, using the quadratic formula provides a more reliable method.
The quadratic formula states that for the equation ax^2 + bx + c = 0, the values of x can be found using:
x = (-b ± √(b^2 - 4ac)) / (2a)
Applying this formula to our equation -16t^2 - 20t + 200 = 0, we have:
t = (-(-20) ± √((-20)^2 - 4*(-16)*200)) / (2*(-16))
Simplifying further, we have:
t = (20 ± √(400 + 12800)) / (-32)
t = (20 ± √13200) / -32
At this point, you can use a calculator to find the approximate values for t. The positive root will give the time when the package is at the highest point in its trajectory, while the negative root will give the time when the package hits the ground.
Once you know the time at which the package hits the ground, you can calculate the speed of the package at impact. The speed is simply the magnitude of the velocity vector at that time. Using the equation v(t) = v0 + at, where v(t) is the velocity at time t and a is the acceleration due to gravity (-32 ft/sec^2), you can find the speed of the package at impact.
I hope this explanation helps you understand the process of solving this problem!
You are correct that you can use the equation -16t^2 + 20t + 200 = 0 to calculate the time it takes for the package to hit the ground. However, factoring might not be the simplest way to solve this equation. Let me guide you through solving it using a different method called the quadratic formula.
The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, the quadratic equation is -16t^2 + 20t + 200 = 0, with a = -16, b = 20, and c = 200.
Using the quadratic formula, we can calculate the time it takes for the package to hit the ground:
t = (-20 ± √(20^2 - 4(-16)(200))) / (2(-16))
Simplifying:
t = (-20 ± √(400 + 12800)) / (-32)
t = (-20 ± √(13200)) / (-32)
t = (-20 ± √(33 × 400)) / (-32)
t = (-20 ± 20√33) / (-32)
t ≈ 6.31 seconds or -0.944 seconds
Since negative time is not meaningful in this context, we can ignore the -0.944 seconds solution.
Therefore, it will take approximately 6.31 seconds for the package to hit the ground.
To determine the speed of the package at impact, we can use the equation v(t) = -32t + 20, which represents the velocity of the package at time t.
Plugging in t = 6.31 into the equation, we get:
v(6.31) = -32(6.31) + 20
v(6.31) = -201.92 + 20
v(6.31) ≈ -181.92 ft/sec
Hence, the speed of the package at impact is approximately 181.92 ft/sec downward.