A helicopter is rising at 5.1 m/s when a bag of its cargo is dropped. (Assume that the positive direction is upward.)
(a) After 2.0 s, what is the bag's velocity?
(b) How far has the bag fallen?
(c) How far below the helicopter is the bag?
part c is what I am having trouble with. what equation do I use?
delta Y (distance of bag below helicopter)
= Yhelicopter - Ybag
= Yo + 5.1 t - (g/2) t^2 - Yo -5.1t
= -(g/2)t^2
To solve this problem, we need to use the equations of motion. In this case, we'll use the equations for vertical motion since the bag is dropped vertically.
(a) After 2.0 s, what is the bag's velocity?
To find the bag's velocity after 2.0 seconds, we can use the equation:
v = u + gt
where:
v = final velocity
u = initial velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
In this case, the bag is dropped, so its initial velocity (u) is 0 m/s. Plugging in the values, we get:
v = 0 + (9.8 m/s^2)(2.0 s)
v = 19.6 m/s
Therefore, after 2.0 seconds, the bag's velocity is 19.6 m/s upwards.
(b) How far has the bag fallen?
To find the distance the bag has fallen, we can use the equation:
s = ut + (1/2)gt^2
where:
s = distance
u = initial velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Since the bag is dropped, its initial velocity (u) is 0 m/s. Plugging in the values, we get:
s = 0 + (0.5)(9.8 m/s^2)(2.0 s)^2
s = 19.6 m
Therefore, the bag has fallen 19.6 meters.
(c) How far below the helicopter is the bag?
Since the helicopter is rising, and the bag is falling, we can calculate the distance between the helicopter and the bag by finding the negative value of the distance the bag has fallen. Therefore, the bag is 19.6 meters below the helicopter.
Use the equations
V = 5.1 - g t
Y = Yo + 5.1 t - (g/2) t^2
Yo is the elevation where the release occurs. You know what g is, I'm sure.
For (c), remember that the helicopter will be 10.2 m higher than Yo at t = 2 s, sincve it continues to rise. The bag will be lower than Yo.