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Introductory Physics Homework Help
How Do You Calculate the Constant in a Helicopter's Takeoff Equation?
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[QUOTE="RJLiberator, post: 4981878, member: 504241"] [h2]Homework Statement [/h2] As a helicopter carrying a crate takes off, its vertical position (as well as the crate's) is given as: y(t)= At^3, where A is a constant and t is time with t=0 corresponding to when it leaves the ground. When the helicopter reaches a height of h = 15.0m the crate is released from the underside of the aircraft. From the time the crate leaves the helicopter to the time it hits the ground, 2.50 seconds pass. Calculate A. [h2]Homework Equations[/h2] Kinematic equation: S = Si + Vi*t+1/2*a*t^2 given equation: y(t)=At^3 [h2]The Attempt at a Solution[/h2] I first found the initial velocity that the crate left the aircraft using kinematics: 0=15+Vi*2.5+1/2*-9.81*2.5^2 Vi = 6.2625 m/s Next, I took the derivative of the position function and got the velocity function y'(t)=3At^2 I then set 6.2625 = 3A*t^2 and noticed that using the first equation A=15/t^3 6.2625=3*15/t^3*t^2 simplified to 6.2625=45/t t=7.1865 I then inputted this t time back into the original equation to find A 15=A*7.1865^3 A=0.0404 I added units of m/s^3 to the answer so that it cleared out in units. Does this seem right? Is it okay to add the units at the end here as m/s^3? I don't like the answer A=0.0404, it seems to small. [/QUOTE]
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Introductory Physics Homework Help
How Do You Calculate the Constant in a Helicopter's Takeoff Equation?
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