A helicopter is ascending vertically with a speed of 4.00 m/s. At a height of 185 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?
The initial velocity of the package is also the velocity of the helicopter, 4.00m/s.
The displacement required to reach the ground is 185m.
Acceleration is due to gravity.
Now use the equation d = Vi * t + (1/2)*a*t^2 to find time.
-185m = 4.00m/s * t - (1/2)* g * t^2
You will need to use the Quadratic formula.
Note how the signs are placed - positive is upwards, negative is downwards.
To find the time it takes for the package to reach the ground, we can use the equation of motion for vertical motion:
h = ut + (1/2)gt^2
where:
h = vertical displacement (in this case, the height of the helicopter from the ground)
u = initial vertical velocity (4.00 m/s)
t = time taken
g = acceleration due to gravity (approximately 9.8 m/s^2)
In this case, the initial height (h) is 185 m, and the package will fall downwards, so the acceleration due to gravity (g) is positive.
Using these values, we can rearrange the equation to solve for time (t):
h = ut + (1/2)gt^2
185 = (4.00)t + (1/2)(9.8)t^2
Simplifying the equation:
(1/2)(9.8)t^2 + 4.00t - 185 = 0
Now, we can solve this quadratic equation for t either by factoring, completing the square, or using the quadratic formula. Once we solve for t, we can determine the time it takes for the package to reach the ground.