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Homework Statement
Besides its weight, a 4.20 kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of (3.80i hat - 3.30j hat) m, where the direction of j hat is the upward vertical direction. Determine the other force.
Homework Equations
F = ma; F =\sqrt{F<sub>x</sub><sup>2</sup>+F<sub>y</sub><sup>2</sup>}
The Attempt at a Solution
okay so i know that the weight is pulling down on the object at 41.16 N. So i need to find the direction and magnitude of a force countering it, so that the resultant vector will = 5.03, however I'm confused about what role 1.20 s has to do with the problem. any help?