Question 10, Physics - equation sheet attached A rock is thrown straight up with initial speed of 15 m/s. At one instant later, the rock is moving at aspeed of 4.3 m/s, where is the rock at that instant?0.54 m above the launch point.11 m above the launch point.0.54 m below the launch point.O 11 m below the launch point.O Information provided is not enough to answer. Physics 114 Equation SheetConstants and ConversionsKinematics Continuedg = 9.80 m/sFree-fall accelerationΔνInstantaneousainst. = limAt-o AtAcceleration1N = 1 kg m/s?NewtonUniform motion(v) = (v); = constantPosition in uniformX = x + (v)AtMathematics, Scaling and Vectorsb = a* + loga (b) = xmotionLogarithmConstant(v); = (v,); + azAtacceleration:1log(ab) = log(a) + log (b)x, = x, + (v,),At +a, (at)?log Ax" = n log x + log A(v,); = (v,)} + 2a,AxVolume of a sphereV =Surface area of aA = 4ar?ForcessphereNewton's second lawFnet = EF = mã%3DVolume of a cylinderV = arlNewton's second lawFnetx = EF = ma,%3DSurface area of aA = 2ar? + 2rlComponent formFnety = ER, = maycylinderMass densityp = m/VNewton's Third LawFA en =-ton AA, = A cos e (rel. to x-axis)Weightw = mgx -component of avector ÅApparent weightWapp = magnitude of supporting forcesy -component of aAy = A sin 8 (rel to x-axis)vector ÅKinetic frictionfk = HanMagnitude of vector ẢStatic frictionA = JA + A,Reynolds numberRe = pvl/nDirection of A relative8 = tan-(Ay/A,)1Drag (high Reynoldsnumber)=CopAv?to x-axisAddition of two vectors If = Å + B, thenC, = A, + B,D = 6nyrvDrag (low Reynoldsnumber)Cy = Ay + ByCircular MotionKinematicsCentripetal accelerationa =DisplacementAx = x - XAverage VelocityAxFrequency1VargT2arAtAxVinst. = limInstantaneous VelocityAt+0 AtAvAverage AccelerationdavgΔε

Name: Question 10, Physics - equation sheet attached A rock is thrown straig
Uploaded: 2024-03-05T11:56:34.000Z
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Description: Question 10, Physics - equation sheet attached A rock is thrown straight up with initial speed of 15 m/s. At one instant later, the rock is moving at aspeed of 4.3 m/s, where is the rock at that instant?0.54 m above the launch point.11 m above the la