The height of a helicopter above the ground is given by h = 2.80t3, where h is i
ID: 2054931 • Letter: T
Question
The height of a helicopter above the ground is given by h = 2.80t3, where h is in meters and t is in seconds. After 2.20 s, the helicopter releases a small mailbag. Assume the upward direction is positive and the downward direction is negative.a.) What is the velocity of the mailbag when it is released?
b.) What maximum height from the ground does the mailbag reach?
c.) What is the velocity of the mailbag when it hits the ground?
d.) How long after its release does the mailbag reach the ground?
Explanation / Answer
a)dh/dt = v = 8.4 t^2 = 8.4 * 2.2^2 = 40.656 m/s , upwards wrt to ground will b)v^2=u^2 - 2gs s above the point of release = 84.33 m height of the helicopter when bag was released = 2.8*2.2^3 = 29.81 m so maximum height = 114.14 m c)velocity of the bag = sqrt(2*9.8*114.14) = 47.29 m/s d)v = u - gt for upward distance 40.656/9.8 = t1 = 4.148 sec for falling down frm maximum height = sqrt(2h/g) = sqrt(2*114.14/9.8) = 4.82 sec total time = 8.974 sec
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