A batter hits a baseball, giving it an initial velocityof 41m/s at 47 degrees ab

Question

A batter hits a baseball, giving it an initial velocityof 41m/s at 47 degrees above the horizontal. It is a home rune, andthe ball is caught by a fan in the stands. The vertical componentof the velocity of the ball when the fan caught it was -11m/s. Howhigh is the fan seated aboive the field ? I tried like 20 times still coundt find it. Help me out xD A batter hits a baseball, giving it an initial velocityof 41m/s at 47 degrees above the horizontal. It is a home rune, andthe ball is caught by a fan in the stands. The vertical componentof the velocity of the ball when the fan caught it was -11m/s. Howhigh is the fan seated aboive the field ? I tried like 20 times still coundt find it. Help me out xD

Explanation / Answer

Find the y component of velocity: v(y) = 41Sin47 v(y) = 29.99 m/s -- With the conservation of energy, we know: 1/2mv^2 = mgh -- In the vertical direction energy at the bottom is equal to theenergy at the maximum reached height: -- solve for h, and this is the maximum height above thefield: h(1) = (0.5)*(29.99^2) / 9.8 ----> m cancels h(1) = 45.89 m -- Now, the velocity of the ball when caught was -11 m/s, soagain: mgh = 1/2mv^2 -- solve for h again: h(2) = (0.5)*(-11^2) / 9.8 ----> m cancels h(2) = 6.17 m -- -- Height above the field: H = h(1) - h(2) H = 45.89 - 6.17 H = 39.7 m ----> fan above field level -- -- Hope this helps. h(2) = (0.5)*(-11^2) / 9.8 ----> m cancels h(2) = 6.17 m -- -- Height above the field: H = h(1) - h(2) H = 45.89 - 6.17 H = 39.7 m ----> fan above field level -- -- Hope this helps.

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