- #1
intriqet
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Homework Statement
The height of a helicopter above the ground is given by h=2.90t^3, where h is in meters and t is in seconds. After 1.80 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 max 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?
Homework Equations
Kinematic equations
The Attempt at a Solution
So for the first part shouldn't the velocity at release be zero? if not what time interval would I use to derive velocity at release?
Also, max height should be 2.90(1.80)^3 = 16.9 m
if part a is zero then the final velocity can be derived by equation Vf^2 - Vi^2 = 2a * deltaX.
= -18.2 m/s
the velocity
However, the online service won't accept my answer. :/