Description
A particle moving at a velocity of 2.3 m in the positive x direction is given an acceleration of 4.9 m/s^2 in the positive y direction for 4.3s. What is the speed of the particle? Answer in units of m/s.
Explanation & Answer
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if a=4.9m/s^2 after 4.3s the velocity will be (Vf=a x t)so 4.9 x 4.3=21.07m/s in y direction and 2.3m/s in x direction
so the total speed is sqrt[(21.07)^2 +(2.3)^2]= 21.2m/s with the angle of arctg (21.07/2.3)= 83 degree