The height of a helicopter above the ground is given by h = 2.55t3, where h is in meters and t is in seconds. At t = 1.60 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?
h = 2.55t^3 = 2.55(1.60)^3 = 10.44 m.
h = Vo*t + 0.5g*t^2 = 10.44 m,
0 + 4.9t^2 = 10.44,
t^2 = 2.13,
t = 1.46 s.
To find the time it takes for the mailbag to reach the ground after it is released, we need to set up an equation and solve it.
We know that the height above the ground is given by the equation h = 2.55t^3, where h is the height in meters and t is the time in seconds.
When the mailbag hits the ground, the height is zero. So we can set up the equation:
0 = 2.55t^3
To find t, we need to solve this equation for t.
Dividing both sides by 2.55, we get:
0 = t^3
Since any number raised to the power of 3 will be zero if and only if the number itself is zero, we have:
t = 0
Therefore, the mailbag reaches the ground immediately after it is released, or at t = 0 seconds.