Best Answer

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.

Q: 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?

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Distance of fall in T seconds = 1/2 g T2Distance of fall in 2 seconds = (1/2) (9.8) (2)2 = (4.9 x 4) = 19.6 metersHeight of this particular ball after 2 seconds = (70 - 19.6) = 50.4 meters

because top hat

In 1 minute it will rotate half as many times = 385. Since is minute is 60 seconds then the rate is 385/60 per second = 6.41666.

5 minutes + 10 seconds = 5 * 60 seconds + 10 seconds = 300 seconds + 10 seconds = 310 seconds

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