A helicopter is rising at 5.5 m/s when a bag of its cargo is dropped. (Assume that the positive direction is upward.)
(a) After 1.5 s, what is the bag's velocity?
(b) How far has the bag fallen?
(a) v = v0 + at, v0 = 5.5 m/s, a = -g = -9.80 m/s.
(b) y = y0 + v0t + (1/2)at2
y-y0 = v0t + (1/2)at2
at2 is acceleration x time squared.
To solve this problem, let's break it down step by step.
Given:
The helicopter is rising at 5.5 m/s (upward velocity)
Time taken = 1.5 s
(a) To find the bag's velocity after 1.5 seconds, we need to consider the effect of gravity on the bag's motion. We can use the following kinematic equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Since the bag is being dropped, its initial velocity is 0 m/s (as it was with the helicopter). The acceleration due to gravity, considering the positive direction is upward, is -9.8 m/s^2 (negative because it acts in the opposite direction).
Substituting the values into the equation:
v = 0 + (-9.8) * 1.5
v = -14.7 m/s
Therefore, the bag's velocity after 1.5 seconds is -14.7 m/s. The negative sign indicates that the bag is moving downwards.
(b) To find how far the bag has fallen, we can use the equation of motion for vertical free fall:
s = ut + (1/2)at^2
Where:
s = distance fallen
u = initial velocity (0 m/s in this case)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (1.5 s)
Substituting the values into the equation:
s = 0 * 1.5 + (1/2) * (-9.8) * (1.5)^2
s = 0 - 11.025
s ≈ -11.03 m
Therefore, the bag has fallen approximately 11.03 meters. The negative sign indicates that it has fallen in the opposite direction of the positive direction (upward).
To find the answers to these questions, we need to use the equations of motion. The motion of the bag can be described using the equation:
y = y0 + v0t + (1/2)at^2
where:
y = final position
y0 = initial position
v0 = initial velocity
t = time
a = acceleration
We know that the initial position of the bag (y0) is at the helicopter's height, which we can assume to be 0 meters. The initial velocity (v0) is the helicopter's rising velocity of 5.5 m/s. The acceleration (a) is due to gravity and is equal to -9.8 m/s^2.
(a) After 1.5 seconds, the bag's velocity can be found by applying the equation:
v = v0 + at
where:
v = final velocity
Substituting the known values:
v = 5.5 m/s + (-9.8 m/s^2) * 1.5 s
Calculating this expression, we find:
v ≈ -9.2 m/s (rounded to one decimal place)
Therefore, after 1.5 seconds, the bag's velocity is approximately -9.2 m/s. The negative sign indicates that the bag is moving downward.
(b) To determine how far the bag has fallen, we need to find the displacement (Δy) using the equation:
Δy = y - y0 = v0t + (1/2)at^2
Substituting the known values:
Δy = 5.5 m/s * 1.5 s + (1/2) * (-9.8 m/s^2) * (1.5 s)^2
Simplifying this expression, we find:
Δy ≈ -7.4 m (rounded to one decimal place)
Therefore, after 1.5 seconds, the bag has fallen approximately 7.4 meters. Again, the negative sign indicates downward motion.