Question #115447

A basketball is shot at 12.0 m/s at a 55.0 degree angle. What is the velocity of the ball 2.16 seconds later?

Expert's answer

*The ball has two projections of speed: *"v_{x}" *and *"v_{y}" .

*At the initial moment:*

"v_{0x}=v_{0}\\times \\cos(\\alpha) \\newline\nv_{0y}=v_{0}\\times \\sin(\\alpha)"

*The x-projection of speed will not change in time. *

*The y-projection:*

"v_{y}=v_{0y}+at=v_{0y}-gt"

*Take *"g=10" :

"v_{x}=v_{0x}=12\\times\\cos(55)\\approx6.88\\newline\nv_{y}=12\\times\\sin(55)-10\\times2.16\\approx-11.76"

*So:*

"v=\\sqrt{v_{x}^2+v_{y}^2}=\\sqrt{47.3344+138.2976}\\approx13.63m\/s"

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