An acceleration of 10m/s² up and an acceleration of 2m/s² horizontal. Thanks.
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kuiperbelt2003 has given correct answer,
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the magnitude of the resultant is found using the pythagorean theorem
resultant = sqrt[10^2+2^2] = 10.2m/s/s
the direction of the new vector is found from tan(theta) = 10/2 or theta = 78.7 deg counterclockwise up from the positive x axis
resultant = sqrt[10^2+2^2] = 10.2m/s/s
the direction of the new vector is found from tan(theta) = 10/2 or theta = 78.7 deg counterclockwise up from the positive x axis
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a = sqrt (10^2 + 2^2) = sqrt(104) <<<<<
theta = tan-1 (10/2) = tan-1(5) <<<<<<,
theta = tan-1 (10/2) = tan-1(5) <<<<<<,