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A bat strikes a 0.145-kgkg baseball. Just before impact, the ball is traveling horizontally to the right at 47.0 m/sm/s ; when it leaves the bat, the ball is traveling to the left at an angle of 29.0 ∘∘ above horizontal with a speed of 59.0 m/sm/s . The ball and bat are in contact for 1.71 msms . Part A Find the horizontal and vertical components of the average force on the ball. Let +x+x be to the right and +y+y be upward Express your answers using three significant figures separated by a comma.

User Aeapen
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Answer:

Step-by-step explanation:

The initial velocity of the baseball is:


v_i^(\to)= 47.0 \ m/s

From velocity of baseball;


v_f ^(\to) = -59 cos 29^0 \hat i + 59 sin \ 29 \hat j \\ \\ \implies -51.60 \hat i + 28.60 \hat j


Average \ force <F^(\to)> = (\Delta P ^(\to ))/(t) \\ \\ \implies (m(v_f^(\to )- v_i ^(\to))/(t) \\ \\ \implies (0.145(-51.60 \hat i +28.60 \hat j -31 \jat i))/(1.71 * 10^(-3)) \\ \\ <F^(\to)> = (11.98 \hat i + 3.857 \hat j )/(1.71* 10^(-3)) \\ \\ <F^(\to)> =7005.85 \hat i + 2255.56\hat j \\ \\ <F^(\to)>= (701 * 10^1 , 226 * 10^1) \ Newton

User Richard Williams
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