A helicopter is ascending vertically with a speed of 5.20m/s. At a height of 125 m above the Earth, a package is dropped from a window. How much time does it take fo the package to reach the ground? [Hint: The package's initial speed equals the helicopter's}
vi= 5.2 m/s
g= -9.8 m/s^2
d=125 m
t= ?
d= vit + gt^2/2
t=5.61s
To find the time it takes for the package to reach the ground, we can use the equations of motion.
First, let's consider the vertical motion of the package. Since the helicopter is ascending vertically with a speed of 5.20 m/s, the initial velocity of the package when it is dropped is also 5.20 m/s.
We know the initial velocity (u) of the package is 5.20 m/s, the final velocity (v) when it reaches the ground is 0 m/s (as it comes to rest), and the distance (s) it travels is 125 m. We want to find the time (t) it takes to reach the ground.
Using the equation of motion for displacement:
s = ut + (1/2)at^2
where
s = displacement (125 m)
u = initial velocity (5.20 m/s)
a = acceleration (acceleration due to gravity, approximately 9.8 m/s^2)
t = time
Plugging in the known values, we get:
125 = (5.20)t + (1/2)(9.8)t^2
Simplifying the equation, we have:
0.5(9.8)t^2 + 5.20t - 125 = 0
This is a quadratic equation in terms of time (t). We can solve it by using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For this equation, a = 0.5(9.8), b = 5.20, and c = -125.
Plugging in these values, we can solve for t using the quadratic formula:
t = (-5.20 ± √(5.20^2 - 4(0.5(9.8))(-125))) / 2(0.5(9.8))
Simplifying further, we get:
t = (-5.20 ± √(27.04 + 245)) / 9.8
t = (-5.20 ± √(272.04)) / 9.8
t = (-5.20 ± 16.49) / 9.8
Now, we have two possible values for t:
t1 = (-5.20 + 16.49) / 9.8 ≈ 1.14 seconds
t2 = (-5.20 - 16.49) / 9.8 ≈ -2.30 seconds
Since time cannot be negative in this context, we disregard the negative value. Therefore, the time it takes for the package to reach the ground is approximately 1.14 seconds.