A helicopter is ascending vertically with a speed of 4.50 m/s. At a height of 195 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.]
hf=0=hi+vi*t+1/2 g t^2
solve this squadratic equation for t
hi=195
g=-9.8
vi=4.50
To solve this problem, we need to consider the motion of the package as it falls from the helicopter. Let's break down the problem step by step:
1. We know that the initial speed of the package is 4.50 m/s, which is equal to the ascending speed of the helicopter. This means that initially, the package is moving upwards relative to the ground, but at some point, it will start falling.
2. When the package is dropped from the helicopter, its initial speed becomes zero. This is because the package is now only affected by the force of gravity and no longer has any upward propulsion. So, the initial velocity (v₀) of the package is 0 m/s.
3. The package will continue to accelerate downwards due to the force of gravity, which is approximately 9.8 m/s² near the surface of the Earth. Therefore, the acceleration (a) of the package is -9.8 m/s² (negative because it's acting in the opposite direction to the initial motion).
4. We can use the kinematic equation to find the time it takes for the package to reach the ground, given the initial velocity, acceleration, and displacement. The kinematic equation we'll use is:
s = v₀t + (1/2)at²
where:
s = displacement (195 m)
v₀ = initial velocity (0 m/s)
t = time (unknown)
a = acceleration (-9.8 m/s²)
5. Plugging in the values into the kinematic equation, we get:
195 = 0*t + (1/2)*(-9.8)*t²
6. Simplifying the equation gives us:
195 = -4.9t²
7. Rearranging the equation to solve for t:
t² = -195 / -4.9
t² = 39.79591837
t ≈ √39.79591837
t ≈ 6.31 seconds (rounded to two decimal places)
Therefore, it takes approximately 6.31 seconds for the package to reach the ground.