A softball is hit over a third baseman\'s head with speed v_0 and at an angle th

Question

A softball is hit over a third baseman's head with speed v_0 and at an angle theta from the horizontal. Immediately after the ball is hit, the third baseman turns around and runs straight back at a constant velocity V = 7.000m/s, for a time t=2.000s. He then catches the ball at the same height at which it left the bat. The third baseman was initially L=18.00m from the location where the ball was hit at home plate.

i have found:
Vo = 18.8
theta = 31.5

0.100s before the ball is caught..
Use the notation vx, vy, an ordered pair of values separated by a commas Express your answer numerically in meters per second to three significant figures
vx=16.0, vy=-8.82

THIS IS MY QUESTION!!! :
Find a vector r expression for the position of the softball 0.100 s before the ball is caught.

Use the notation x, y, an ordered pair of values separated by a comma, where x and y are expressed numerically in meters, as measured from the point where the softball initially left the bat. Express your answer to three significant figures.

Explanation / Answer

Let the x axis be the line from home through third base Let the height above bat be Z First find how far the ball tralled in x direction. X = Xo + V t X = 18 + 7 * 2 X = 32 m from home plate t = 2 secs Vx = 32 / 2 m/s 16 m/s Know consider component of velocity in vertical direction Z = Zo + Vzo*t + 0.5 * Az * t^2 Z = 0 + Vzo*t + 0.5* -9.8 * t^2 But when the ball is caught Z = 0 Vzo*t + 0.5* -9.8 * t^2 = 0 t(Vzo - 4.9*t) = 0 this expression is only equal to zero when : t = 0 The time of the batter hitting the ball Vzo - 4.9t = 0 but we know the ball was in the air for two secs Vz0 4.9*2 Vz0 = 9.8 m/s In vector notation (do you know this? Its very cool. I is a vector one unit long in the x direction, k is a vector one unit long in the z direction, cf Rec notation and Polar notation)) Vo = 16 * I + 9.8 * k |Vo| = sqrt(16^2 + 9.8^2) |Vo| = 18.7 m/s Part 2. Theta in degrees Tan(@) = 9.8/16 0.6125 @ ATAN(9.8/16) @ 0.549 rad @ .555*180/pi() @ 31.8 degrees 3 Can I use vector notation now I have explained it? Vx = 16 Vz = 9.8 - 9.8*t So V = 16* i_ + (9.8 - 9.8*t) * k_ So 0.1 secs before catch t = 1.9secs So substitute into your velocity equation Vx = 16 Vz = 9.8 - 9.8*1.9 Vz = -8.82 m/s Looks good - not quite -9.8 m/s = speed at catching) Convert into polar notation V = sqrt(16^2+(-8.82)^2) V = 18.3 m/s @ atan(-8.82/16) -28.9 degrees Looks good not quite but almost the launch angle 4 X(t) = 16*t Z(t) = Voz * t - 0.5*9.8*t^2 Rvec = 16*t*i_ + (Voz * t - 0.5*9.8*t^2)*k_ Substitute in t = 1.9secs X = 30.4 m Almost but not quite 32m Z = 0.931 m Almost but not quite zero Polar notation r = sqrt(30.4^2+0.931^2) r = 30.4142526 @ Atan(0.931/30.4) Almost but not quite 32m @ 1.75 degrees Almost but not quite zero

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