A helicopter is ascending vertically with a speed of 5.50 m/s. At a height of 155 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [Hint: The package's initial speed equals the helicopter's.]
To determine how much time it takes for the package to reach the ground, we can use the equation of motion for vertical motion:
\[ h = v_i t + \frac{1}{2}gt^2 \]
Where:
- \( h \) is the height of the package above the ground (155 m)
- \( v_i \) is the initial velocity of the package (equal to the helicopter's velocity, so 5.50 m/s)
- \( g \) is the acceleration due to gravity (-9.81 m/s²)
- \( t \) is the time it takes for the package to reach the ground (what we want to find)
Rearranging the equation, we get:
\[ 0.5gt^2 + v_i t - h = 0 \]
Now, we can solve this quadratic equation to find the value of \( t \). We can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, the values are:
- \( a = 0.5g \) (0.5 multiplied by the acceleration due to gravity)
- \( b = v_i \) (the initial velocity of the package)
- \( c = -h \) (the negative height)
Substituting these values, we get:
\[ t = \frac{-v_i \pm \sqrt{v_i^2 - 4(0.5g)(-h)}}{2(0.5g)} \]
Now, let's calculate the value of \( t \):
\[ t = \frac{-5.50 \pm \sqrt{5.50^2 - 4(0.5)(-9.81)(-155)}}{2(0.5)(-9.81)} \]
Simplifying the expression:
\[ t = \frac{-5.50 \pm \sqrt{30.25 + 3096.08}}{-9.81} \]
\[ t = \frac{-5.50 \pm \sqrt{3126.33}}{-9.81} \]
\[ t = \frac{-5.50 \pm 55.92}{-9.81} \]
Now we have two possible values for \( t \):
1. \[ t_1 = \frac{-5.50 + 55.92}{-9.81} \]
2. \[ t_2 = \frac{-5.50 - 55.92}{-9.81} \]
Calculating \( t_1 \):
\[ t_1 = \frac{50.42}{-9.81} \approx -5.14 \; \textrm{(rejected as time cannot be negative)} \]
Calculating \( t_2 \):
\[ t_2 = \frac{-61.42}{-9.81} \approx 6.26 \]
Therefore, it takes approximately 6.26 seconds for the package to reach the ground.
To find the time it takes for the package to reach the ground, we can use the formula for motion with constant acceleration:
h = (1/2)gt^2
Where:
h = height (155 m)
g = acceleration due to gravity (-9.8 m/s^2, assuming no air resistance)
t = time
Since the package is dropped from rest (initial speed equals 0), we can rewrite the equation as:
h = (1/2)gt^2 + v₀t
Where:
v₀ = initial speed of the package (equal to the speed of the helicopter, 5.50 m/s)
In this case, the helicopter is ascending, which means the initial speed of the package is upward (positive). Therefore, we will take the positive value for v₀.
Plugging in the values into the equation:
155 = (1/2) (-9.8) t^2 + 5.50 t
Rearranging the equation to bring all terms to one side, we get:
(1/2) (-9.8) t^2 + 5.50 t - 155 = 0
Now we can solve this quadratic equation to find the value of t.