hello, I need some help with this work problem that goes... If the height of a helicopter above the ground is given by h=3.25t^3, where h is in meters and t is in sec. After 1.50s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Question

hello, I need some help with this work problem that goes... If the height of a helicopter above the ground is given by h=3.25t^3, where h is in meters and t is in sec. After 1.50s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

So, to approach this question, I first 1.50s in the formula for h given to get my height in which I got 10.97m... then i am using the formula x=v(i)t-1/2gt^2 where my x is -10.97, the v is 0 and g=8.91 when a substitute all my values to the equation i and up with 2.24=t^2 in which then i square root both sides to give and answer of about 1.49s... I thought that my calculation were right but it seems that the answer is incorrect... please let me know what i am doing wrong... thank you

Answer 1

You need the velocity of the helicopter when the bag is dropped.

h=3.25t^3
dh/dt= 9.75 t^2= vi

h(1.5)=3.25*1.5^3=10.97

Now, for the bag:
hf=hi+vit-4.9t^2
0=10.97+9.75(1.5^2)t-4.9t^2

Now, solve that quadratic.

Answer 2

Hi! It looks like you're on the right track, but there might be a small mistake in your calculation. Let's go through the steps again to make sure we get the correct answer.

First, we find the height of the helicopter after 1.50 seconds by substituting t = 1.50 into the equation h = 3.25t^3:

h = 3.25(1.50)^3
h ≈ 10.97 meters

So you calculated this correctly.

Next, we need to find the time it takes for the mailbag to reach the ground after it is released. We can use the second equation you mentioned: x = v(i)t - 1/2gt^2.

Let's break down the equation:

x: The height/displacement, which is -10.97 meters since it's moving downwards.

v(i): The initial velocity. Since the mailbag is released, the initial velocity is 0 m/s.

g: The acceleration due to gravity, which is approximately 9.81 m/s^2.

t: The time we want to find.

So we have:

-10.97 = 0t - 1/2 * 9.81 * t^2

Now, let's solve for t. Rearrange the equation:

0 = 1/2 * 9.81 * t^2 - 10.97

Multiply both sides by 2 to remove the fraction:

0 = 9.81 * t^2 - 21.94

Rearrange the equation again:

9.81 * t^2 = 21.94

Divide both sides by 9.81:

t^2 ≈ 2.24

Now take the square root of both sides to solve for t:

t ≈ √2.24
t ≈ 1.50 seconds (approximately)

So it looks like you made an error in the last step of your calculation. The correct answer is approximately 1.50 seconds, not 1.49 seconds.

I hope this clears up any confusion! Let me know if you have any further questions.