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The height of a helicopter above the ground is given by h 3.00t3 where h is in meters and t is in seconds After 2.00s the helicopter releases a small mailbag How long after its release does the mai?
Wiki User
∙ 12y ago
Height of the helicopter above ground = 3 T3 . After 2 seconds, H = 24 meters .
Speed of ascent = 9 T2 . After 2 seconds, it's rising at 36 meters per second.
So the mailbag is released from a height of 24 meters, with an upward velocity of 36 m/s.
The height of the mailbag after its release is H = H0 + V0T + 1/2 g T2
H = height at any time after release = 0 when it hits the ground
H0 = height when released = + 24 meters
V0 = velocity when released = 36 m/s up = +36 m/s
g = grav acceleration = 9.8 m/s2 down = -9.8 m/s2
T = time from release until it hits the ground
H = H0 + V0T + 1/2 g T2
0 = 24 + 36 T - 4.9 T2
4.9 T2 - 36 T - 24 = 0
T = 1/9.8 [ 36 +/- sqrt(1296 + 470.4) ] = 1/9.8 [ 36 +/- 42.029 ]
T = -0.6152 sec
T = 7.962 sec after release.
Wiki User
∙ 12y ago
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