A particle moves along a path y=x^2+4x+100, starting with an initial velocity of Vo=(4i-16j) m/s. If Vx is constant, what is value of Vy and ay at x=16m?
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4
Explanation:
V= dy/dt = 2x+4
x=16m
2*16+4= 36
Vy = 36*4= 144m/s
here 4 is the value of velocity in x direction.
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