A rescue plane is flying horizontally at a height of 132 m above the ground. The pilot spots a survivor and releases an emergency kit with a parachute. The kit descends at a constant vertical acceleration of -6.89 m/s^2 and the initial plane horizontal speed is 68.9 m/s. Assuming air resistance and wind are negligible, how long will the emergency kit be falling through the air?
To find the time it takes for the emergency kit to fall through the air, we can use the kinematic equation for vertical motion:
Δy = v0*t + 0.5*a*t^2
Where:
Δy = change in vertical distance (height above the ground)
v0 = initial vertical velocity (0 m/s since the kit is released with no initial velocity in the vertical direction)
a = vertical acceleration (-6.89 m/s^2, negative because the acceleration is in the opposite direction of the positive vertical direction)
t = time
Given:
Δy = 132 m
a = -6.89 m/s^2
Plugging in these values into the equation, we get:
132 = 0*t + 0.5*(-6.89)*t^2
Simplifying the equation, we get:
66 = -3.445*t^2
Dividing both sides by -3.445, we get:
t^2 ≈ -19.157
Since time cannot be negative, we discard the negative solution. Therefore, we have:
t ≈ √(19.157) ≈ 4.38 s
So, the emergency kit will be falling through the air for approximately 4.38 seconds.
To find the time it takes for the emergency kit to fall through the air, we can use the equation of motion for vertical motion:
h = ut + (1/2)at^2
Where:
h = height = 132 m (given)
u = initial vertical velocity = 0 m/s (since the kit starts from rest)
a = vertical acceleration = -6.89 m/s^2 (given)
t = time (unknown)
Plugging in the values into the equation:
132 = 0t + (1/2)(-6.89)t^2
264 = -6.89t^2
Rearranging the equation:
t^2 = -264 / -6.89
t^2 = 38.26
t ≈ √38.26 ≈ 6.19 s
Therefore, it will take approximately 6.19 seconds for the emergency kit to fall through the air.
distance = 1/2 * acceleration * time^2
-132 = 1/2 * -6.89 * t^2