Initially, a particle is moving at 4.20m/s at an angle of 34.5∘ above the horizontal. Two seconds later, its velocity is 6.50m/s at an angle of 50.0∘ below the horizontal.



What was the particle's average acceleration during these 2.00 seconds?

1 answer

change in x speed = 6.5 cos50-4.2cos34.5

change in y speed = -6.5 sin50-4.2sin34.5

ax = change in x speed/2
ay = change in y speed/2

a = sqrt (ax^2+ay^2)

A = angle of a above horizontal
tan A = ay/ax