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A soccer player hits a ball with a velocity of 22 m/s at an angle of 36.9 above the horizontal. Air resistance can be ignored. a. b. c. (3) What are the x and y components of the balls initial velocity? (3) How high does the ball go? (4) How long (time) does it take to get to the maximum height?

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prosborough198p2ub41
Answers:

a. Vx=17.6m/s
Vy=13.2m/s

b. 8.90m

c. 1.80s




Vi=22m/s

Theta=36.9deg

Vx=Vi*cos(theta)

Vy=Vi*sin(theta)

h(Max)=Vi^2*sin(theta)^2/2g

t(max height)= Vy/g

Vx=22*cos(36.9)=17.6m/s

Vy=22*sin(36.9)=13.2m/s

h(Max)=22^2*sin(36.9)^2/2(9.8)=8.90m

t(Max height)=17.6/9.8=1.80s

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