A helicopter is ascending vertically with a speed of 5.66m/s. At a height of 106m above the Earth, a package is dropped from the helicopter. How much time does it take the package the package to reach the ground? (Ignore air resistance)
I would like to know what my givens are and what formula(s) to use. Thank you.
h = 4.9t^2 - 5.66t = 106
h = 4.9t^2 -5.66t - 106 = 0
Use Quadratic Formula.
t = 5.264 s.
To solve this problem, you need to determine the time it takes for the package to reach the ground when dropped from a certain height.
Given:
- Initial speed of the helicopter (v₀) = 5.66 m/s
- Height above the ground from which the package is dropped (h) = 106 m
- Acceleration due to gravity (g) = 9.8 m/s² (Earth's standard gravity)
Formula:
To solve this problem, you can use the kinematic equation for vertical motion:
h = v₀t + (1/2)gt²
Where:
- h is the height
- v₀ is the initial speed
- t is the time
- g is the acceleration due to gravity
In this case, the final height (h) is 0 since the package reaches the ground. The initial speed (v₀) is effectively 0 since the package is only dropped, and the acceleration due to gravity (g) is acting downwards.
Plugging in the values, our equation becomes:
0 = (1/2)gt²
Simplifying, we get:
t² = (2h) / g
Solving for t, we take the square root of both sides:
t = √((2h) / g)
Substituting the given values:
t = √((2 * 106) / 9.8)
Now you can use a calculator to find the square root of ((2 * 106) / 9.8), and that will give you the time it takes for the package to reach the ground.