Helicopter Problem: Dropping an Inflatable Raft

[Rowie25]

Homework Statement

A helicopter drops an inflatable raft to flood victims floating in a lake. When the raft is released the helicopter is 50 m directly above the victims and flying at a velocity of 10 m/s at an angle of 30 degrees below the horizontal. Neglect air resistance.
a. How long is the package in the air?
b. How far from the victims does the raft land?
c. Assuming the helicopter drops the package at coordinates (0,0). If the helicopter continues to fly at a constant velocity, where is the helicopter when the package lands?



I think the fact that the helicopter is moving is confusing me. Wouldn't you use the formula y=y_o+v_oysin(angle)(t)-(1/2)gt^2 ?? So you find the time? And then you could use v_y=v_oy-gt to find the velocity? I am not sure how to find c. Please help!

 

[jbsteele]

Homework Equationsy=y_o+v_oysin(angle)(t)-(1/2)gt^2 v_y=v_oy-gt The Attempt at a Solution To find a, use the equation y=y_o+v_oysin(angle)(t)-(1/2)gt^2 50=0+10sin30(t)-(1/2)(-9.8)t^2 t=1.86 s To find b, use the equation x=x_o+v_oxcos(angle)t x=0+10cos30t x=15.8 m To find c, I am confused.

 

[kerimek]

I can assist in answering the questions posed in this helicopter problem.
a. To find the time the package is in the air, we can use the formula for vertical displacement: y = y0 + v0yt - 1/2gt^2. Since the initial vertical position (y0) is 50m and the initial vertical velocity (v0y) is 10m/s, we can plug in these values and solve for t. This gives us a time of approximately 3.19 seconds.
b. To find the distance from the victims that the raft lands, we can use the formula for horizontal displacement: x = x0 + v0xt. Since the initial horizontal position (x0) is 0m and the initial horizontal velocity (v0x) is 10m/s, we can plug in these values and solve for t. This gives us a distance of approximately 31.9 meters.
c. To find the position of the helicopter when the package lands, we can use the same formula for horizontal displacement and solve for t. This time, we will use the distance from part b as our x value and the velocity of the helicopter (10m/s) as our v0x. This gives us a time of approximately 3.19 seconds. We can then use this time in the formula for vertical displacement to find the vertical position of the helicopter at this time. This will give us a position of (31.9m, 25m) relative to the initial position of (0,0).
I hope this helps in understanding how to approach and solve this problem.