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Question
A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of s(t) = 160t - 16t^2 after t seconds.
how fast is the rock traveling when it is 256 feet above the ground?
what is tha maximum height of the rock?
s(t)=160t-16t^2 is the height as function of time . The velocity as function of time is s'(t):
s'(t)=160-32t . We need to know at which to the rocket reached s=256 . Therefore , we solve
160t-16t^2=256 .. The solutions are to=2 %26 t=8 . We can conclude that t=8 is attached to
the falling of the rocket !
Therefore : S'(to=2)=160-32*2=96 ft/sec .
To find max height , we solve s'(t)=0 :
160-32t=0 --> t=5 sec . s(5)=160*5-16*25=370 ft .
Alon.
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Answers by Expert:
Alon Mandes
Expertise
Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
Experience
1. I'm a team member of mathnerds (math site for answering questions)
2. I'm a team member in the Student's Union of the Technion, helping
students who have problems in mathematics.
3. 2 years of experience as a math teacher in college.
4. I give free homework help for high school students in
Mathematics & Physics.
5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" ,
"Complex Functions".
Organizations
Hi-Tech company : GSM4VOIP ; job possition : Algorythm developer.
Education/Credentials
M.A in Mathematics & Bs.c in Electronics.
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