A helicopter is ascending vertically with a speed of 5.20 m/s. AT the height of 125m above the earth, a package is dropped from a window.How much time does it take for the package to reach the ground.
****Note: Please show the steps.
If release occurs at t=0, the vertical height ot the package after that, in meters, is
y = 125 + 5.2 t -4.9 t^2
(The 4.9 is g/2, and g is the acceleration of gravity in units of m/s^2)
Solve the equation y = 0 and you will get the time that it his the ground. It is a quadratic equation with two roots. Take the positive one.
t = [-5.2 -sqrt(5.2^2 +4*125*4.9)]/(-9.8)
= 5.61 seconds
thank you so much!
You a Real One
Why did the package decide to take the express route down? It seems like it wanted to cut through all the traffic on the way. It's always in a hurry!
But okay, let's calculate the time it takes for the package to reach the ground.
We'll start by figuring out how much time it takes for the package to fall from a height of 125m. We can use the equation:
h = (1/2) * g * t^2
Where:
h = height (125m in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Rearranging the equation, we have:
t^2 = (2 * h) / g
t^2 = (2 * 125) / 9.8
Now let's solve for t.
t^2 = 254 / 9.8
t^2 ≈ 25.918
t ≈ √25.918
t ≈ 5.09 seconds
So, it takes the package approximately 5.09 seconds to reach the ground. Just enough time for it to say, "Wheee!" on the way down!
To find the time it takes for the package to reach the ground, we can use the kinematic equation for vertical motion:
h = ut + (1/2) * gt²
where:
h is the height of the helicopter (125m in this case)
u is the initial vertical velocity of the package (0m/s since it's dropped)
g is the acceleration due to gravity (-9.8m/s²)
t is the time taken
Since the package is dropped, the initial vertical velocity u is 0m/s. Therefore, we can simplify the equation to:
h = (1/2) * gt²
Plugging in the known values:
125 = (1/2) * (-9.8) * t²
Let's solve for t:
125 = (-4.9) * t²
Divide both sides by (-4.9):
125 / (-4.9) = t²
-25.51 = t²
Since time cannot be negative, we discard the negative solution. Therefore, we take the square root of the positive value:
t = √25.51
t ≈ 5.05 seconds
Therefore, it takes approximately 5.05 seconds for the package to reach the ground.
A helicopter is ascending vertically with a speed of 5.10 m/s. At a height of 105 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: v0 for the package equals the speed of the helicopter.
Note: Please show the steps.