How Long Does It Take for a Released Mailbag to Hit the Ground?

[ryno2107]

The height of a helicopter above the ground is given by h = 3.50t^3, where h is in meters and t is in seconds. After 1.85 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? (Ans. in s.)

(1) y-y0=v0t+1/2*gt^2

(2) v^2=v0^2+2g(y-y0)

(3) y-y0=1/2*(v0+v)t

I sub. 1.85s for t in h=3.50t^3 and h=y0=22.2m.. This is the initial position. The final position is y=0. So I substituted all of my known values into (1) and got v0= -2.92m/s(initial velocity)
then I sub. all the known values into (2) and got v= -21.1m/s. Finally, I sub. v, v0, and the rest of the known values into (3) and got 1.85s. I believe it is the wrong answer. Please help me to solve this problem.

 

[Delphi51]

I agree with your initial height.
But you can't use a constant acceleration formula to find the velocity of the helicopter at time 1.85 s. There are no formulas for height increasing as the cube of time. If you know a bit of calculus, you can differentiate to get a formula for velocity. If not, you'll have to carefully graph y = 3.50t^3 and draw a tangent line at time 1.85 to estimate the slope, which is the velocity.

 

[tomcue]

Based on the given information, it seems that the mailbag is released from the helicopter at a height of 22.2 meters and an initial velocity of -2.92 m/s. This means that the mailbag is initially moving downwards at a speed of 2.92 m/s. Using this information, we can use the equation (1) to calculate the time it takes for the mailbag to reach the ground.

y-y0=v0t+1/2*gt^2

Where:
y = final position (0 m)
y0 = initial position (22.2 m)
v0 = initial velocity (-2.92 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time (unknown)

Substituting the known values, we get:

0-22.2 = (-2.92)t + 1/2*(-9.8)t^2

Simplifying, we get:

-22.2 = -2.92t - 4.9t^2

Rearranging to get a quadratic equation in standard form:

4.9t^2 + 2.92t - 22.2 = 0

Using the quadratic formula, we get two possible solutions:

t = (-2.92 ± √(2.92^2 - 4*4.9*(-22.2)))/2*4.9
t ≈ 2.32 s or t ≈ -1.55 s

Since the time cannot be negative, we can discard the second solution and conclude that the mailbag reaches the ground after approximately 2.32 seconds after its release from the helicopter.