A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was realeased?
The time to V = 0 derives from Vf = Vo - gt or 0 = 15 - 9.8t yielding t(up) = 1.53sec. resulting in t(dwn) = 16 - 1.53 = 14.47 sec.
h(dwn) then derives from h = Vo(t) + g(t^2)/2 or h = 0 + 4.9t(dwn)^2 yielding h(dwn) = 4.9(14.47)^2 = 1025.96 met.
h(up) then derives from h(up) = 15(1.53) - 4.9(14.47)^2 = 11.48 met.
The vertical rise from ejection to zero vertical velocity derives from h(up) = 15(1.53) - 4.9(1.53)^2 = 11.48 met.
Therefore, the altitude of release is h(rel) = 1025.96 - 11.48 = 1014.48 met.
gasagwa
Well, well, well, looks like we have a real-life skydiving package! Let's figure out how high it was when it took the ultimate plunge.
First, we need to determine the initial vertical velocity of the package when it was dropped. Since the helicopter was moving upward at 15 m/s, we can consider this as the initial velocity. But hang on tight, we're not done yet!
Next, we'll have to calculate the time it took for the package to hit the ground, which is given as 16.0 s. It must have felt like an eternity for that poor package to fall!
Now, let's bring in Newton's second equation of motion into the mix: distance equals initial velocity multiplied by time, plus one-half the acceleration multiplied by the time squared.
In this case, we can assume the package is only influenced by gravity, which means we'll consider the acceleration as -9.8 m/s², due to it going downward.
By plugging in the values, we get:
distance = (15 m/s)(16.0 s) - (1/2)(9.8 m/s²)(16.0 s)²
After evaluating that, we find that the package was released from a height of approximately 940 meters above the ground. Imagine the views it must have had on the way down!
Remember, try not to drop your electrifying packages too often. Safety first!
To find the height above the ground when the package was released, we can use the kinematic equation:
h = h₀ + v₀t - (1/2)gt²
where:
h = height above the ground when the package was released
h₀ = initial height above the ground (unknown)
v₀ = initial velocity of the package (15 m/s)
t = time taken for the package to strike the ground (16.0 s)
g = acceleration due to gravity (approximately 9.8 m/s²)
Let's substitute the known values into the equation and solve for h₀:
h = h₀ + (15 m/s) * (16.0 s) - (1/2) * (9.8 m/s²) * (16.0 s)²
First, let's square the time:
h = h₀ + (15 m/s) * (16.0 s) - (1/2) * (9.8 m/s²) * (256.0 s²)
Now, let's multiply and divide:
h = h₀ + 240 m - (124.8 m/s²) * (256.0 s²)
Next, let's simplify:
h = h₀ - 124.8 m/s² * 256.0 s² + 240 m
h = h₀ - 31948.8 m + 240 m
h = h₀ - 31708.8 m
Finally, let's isolate h₀:
h₀ = h + 31708.8 m
Since we want to find the height above the ground when the package was released, we need to substitute the value of h as 0 (since it's on the ground):
h₀ = 0 + 31708.8 m
h₀ ≈ 31708.8 m
Therefore, the package was approximately 31,708.8 meters (or 31.7 kilometers) above the ground when it was released.
To determine the height at which the package was released, we can use the kinematic equation for displacement:
d = v₀t + (1/2)at²
Where:
d is the displacement (height)
v₀ is the initial velocity
t is the time of flight (time taken to reach the ground)
a is the acceleration due to gravity (which is approximately 9.8 m/s²)
In this case, the package is dropped, so the initial velocity v₀ is 0 m/s, as it is not moving horizontally. The acceleration due to gravity is acting downward, so we will consider it negative (-9.8 m/s²). The time of flight, t, is given as 16.0 s.
Plugging these values into the equation:
d = (0)m/s * (16.0)s + (1/2)(-9.8)m/s² * (16.0)s²
d = 0 + (-4.9 m/s² * 256.0 s²)
d = -1254.4 m
The negative sign in the result indicates that the height is below the reference point (the ground). To find the actual height above the ground, we can take the absolute value of the result:
|d| = 1254.4 m
Therefore, the package was released at a height of approximately 1254.4 meters above the ground.