Asked by Anonymous
A baseball is thrown with a speed of 40m/s at an angle of 30 degrees. Neglecting
friction what is the speed of the ball in m/s when it reaches a height of 10 meters?
friction what is the speed of the ball in m/s when it reaches a height of 10 meters?
Answers
Answered by
Anonymous
Vy = sqrt(Uy^2 - 2gh); Uy = U sin(theta)
Vx = Ux = U cos(theta)
So |V| = sqrt(Vy^2 + Vx^2) = sqrt(Uy^2 - 2gh + Ux^2) = sqrt(U^2 - 2gh) = sqrt(40^2 - 2*9.8*10) = 37.46998799 = 37.47 ANS B.
Note that U^2 = Uy^2 + Ux^2.
Vx = Ux = U cos(theta)
So |V| = sqrt(Vy^2 + Vx^2) = sqrt(Uy^2 - 2gh + Ux^2) = sqrt(U^2 - 2gh) = sqrt(40^2 - 2*9.8*10) = 37.46998799 = 37.47 ANS B.
Note that U^2 = Uy^2 + Ux^2.
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