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I think I got 7.67
v(t)=-t^2+4 -t^2+4t+c -3^2+4(3)+4 7 -0^2+4(0) 0 7
v=dx/dt
integrate v w.r.t.time to find the equation of x
Displacements= -21 ft which means the object traveled in the opposite direction at which it was projected :)
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@ganeshie8
\[s=\int\limits_{0}^{3}(-t^2 +4) dt\]
\[s=(-t^3/3 +4t)_{0}^{3}\]
ohh 3?
let hum/her do the calculations please
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him*
i was talking to @ijlal :)
@rvc was just trying to point out the mistake that she did as included the contact ' C ' ina definite integral didn't post the direct answer though anyways i am sorry :)
na na i reminded u :) what u were doing is alright buddy
@rvc alrighty then :)
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thank you !!
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