For some reason I keep getting this problem wrong. Please Help!
A helicopter is ascending vertically with a speed of 9.00 m/s. At a height of 90 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?
You are headed in the right direction.
That is a distance formula when there is uniform acceleration which is gravity in this problem. The formula is:
s2 = s1 + vt + (1/2)(a)t^2
where:
s2 = final distance
s1 = initial distance
v = initial velocity
a = acceleration
t = total time
Your initial distance above the ground is 90m. The initial velocity will be the vertical velocity of the helicopter when the package is dropped. The final distance will be 0 as you said. Plug all the numbers into the equation and solve for t. Remember that the acceleration will be negative here.
How are you solving it? What answer are you getting?
Well I saw on a related problem someone wrote:
Package height Y at drop time t is
Y = 120 + 5 t - 4.9 t^2
Set Y = 0 and solve.
but I have no idea what that means...
To find the time it takes for the package to reach the ground, we can use the equations of motion. The motion of the package can be divided into two parts: the horizontal motion due to the helicopter's velocity, and the vertical motion due to gravity.
Let's first analyze the vertical motion of the package. The package is dropped from a height of 90 m, so we need to find the time it takes for it to fall this distance.
We can use the equation:
h = 1/2 * g * t^2
where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
Rearranging the equation to solve for time (t), we get:
t^2 = (2h) / g
t^2 = (2 * 90) / 9.8
t^2 = 18.367
Taking the square root of both sides, we find:
t = √18.367
t ≈ 4.287 seconds
Therefore, it takes approximately 4.287 seconds for the package to reach the ground.