The height of a helicopter above the ground is given by h = 2.90t3, where h is in meters and t is in seconds. At t = 1.75 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?
Using standard equation the height is given by:
s=uT+0.5aT^2
if it is dropped u=0
s=0.5aT^2
2.90 t^3 = 0.5 9.8 T^2
substitute value for t as 1.75 s
and find T
tried your way and it didn't work,
To find out how long after the mailbag is released does it reach the ground, we need to find the time at which the height of the mailbag is equal to zero.
Given that the height of the helicopter above the ground is given by the equation h = 2.90t^3, we can substitute h with zero to find the time at which the mailbag reaches the ground.
0 = 2.90t^3
To solve for t, we can rearrange the equation to isolate t:
t^3 = 0 / 2.90
t^3 = 0
Since any number raised to the power of 0 is equal to 1, the equation becomes:
t = ∛0
t = 0
Therefore, the mailbag reaches the ground immediately after it is released, at t = 0 seconds.