A ball is dropped from a helicopter which is descending at 1.5 m/s. After 2 seconds, what are
a) the the velocity of the ball, and
b) how far is the ball below the helicopter?
c) & d) Repeat your calculations for a) and b) if the helicopter had been rising at 1.5 m/s instead.
a. V = Vo + g*t. Vo=1.5 m/s, g=9.8m/s^2
b. d = Vo*t + 0.5g*t^2.
c. Vo = -1.5 m/s,
d. Vo = -1.5 m/s.
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To find the velocity and displacement of the ball in both scenarios, we can use the equations of motion.
a) When the ball is dropped from a descending helicopter:
The initial velocity of the ball is the velocity of the helicopter, which is -1.5 m/s since it is descending (negative direction). The acceleration due to gravity, denoted as "g," is 9.8 m/s^2. The time of descent is given as 2 seconds.
To find the velocity of the ball, we can use the equation:
Final velocity (v) = Initial velocity (u) + (acceleration (a) * time (t))
v = -1.5 m/s + (9.8 m/s^2 * 2 s)
v = -1.5 m/s + 19.6 m/s
v = 18.1 m/s
Therefore, the velocity of the ball after 2 seconds, when dropped from a descending helicopter, is 18.1 m/s.
b) To calculate the distance the ball is below the helicopter, we need to find the displacement.
Displacement (s) = Initial velocity (u) * time (t) + (0.5 * acceleration (a) * time (t)^2)
s = -1.5 m/s * 2 s + (0.5 * 9.8 m/s^2 * (2 s)^2 )
s = -3 m + (0.5 * 9.8 m/s^2 * 4 s^2)
s = -3 m + 19.6 m
s = 16.6 m
Therefore, the ball is 16.6 meters below the helicopter after 2 seconds of being dropped from a descending helicopter.
c) When the helicopter is rising at 1.5 m/s:
In this scenario, the initial velocity of the ball is the velocity of the helicopter, which is 1.5 m/s (positive direction). The acceleration due to gravity remains the same (-9.8 m/s^2).
Using the same equation for finding the velocity:
v = 1.5 m/s + (-9.8 m/s^2 * 2 s)
v = 1.5 m/s - 19.6 m/s
v = -18.1 m/s
Therefore, the velocity of the ball after 2 seconds, when dropped from a rising helicopter, is -18.1 m/s.
d) To find the distance the ball is below the helicopter:
s = 1.5 m/s * 2 s + (0.5 * -9.8 m/s^2 * (2 s)^2 )
s = 3 m + (0.5 * -9.8 m/s^2 * 4 s^2)
s = 3 m - 19.6 m
s = -16.6 m
In this case, the ball would be 16.6 meters above the helicopter after 2 seconds, since the helicopter is rising.
So, the answers are:
a) When the helicopter is descending, the velocity of the ball is 18.1 m/s.
b) The ball is 16.6 meters below the helicopter when dropped from a descending helicopter.
c) When the helicopter is rising, the velocity of the ball is -18.1 m/s.
d) The ball is 16.6 meters above the helicopter when dropped from a rising helicopter.