The height of a helicopter above the ground is given by h = 2.90t3, where h is in meters and t is in seconds. After 2.35 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?
To find the time it takes for the mailbag to reach the ground, we need to set the height equation equal to zero since the ground is at zero height. So we have:
0 = 2.90t^3
Now we can solve for t by taking the cube root of both sides:
t^3 = 0 / 2.90
t^3 = 0
Since any number raised to the power of 3 is still 0, the value of t can be anything. Therefore, the mailbag reaches the ground instantaneously after it is released.
To find out how long after its release the mailbag reaches the ground, we need to determine when the height of the mailbag is zero. We can do this by setting the equation for height, h, to zero and solving for t.
Given: h = 2.90t^3 and t = 2.35 s
We can substitute the value of t into the equation for h and solve for h:
h = 2.90 * (2.35)^3
h = 2.90 * 13.4
h = 38.86 meters
Since the height of the mailbag is the same as the ground when it reaches the ground, we can set h to zero and solve for t:
0 = 2.90t^3
To solve this equation, we need to find the cube root of both sides:
∛0 = ∛(2.90t^3)
Simplifying,
0 = 2.90t
Since anything multiplied by 0 is 0, one solution is t = 0. However, this is not the relevant solution in this case because we are interested in the time after the mailbag is released. So, we can ignore the solution t = 0.
Now, we can solve for t by dividing both sides of the equation by 2.90:
0/2.90 = 2.90t/2.90
0 = t
Therefore, the mailbag reaches the ground at t = 0 seconds after its release.