A softball is hit over a third baseman\'s head with speed v 0 and at an angle fr

Question

A softball is hit over a third baseman's head with speed v0 and at an angle from the horizontal. Immediately after the ball is hit, the third baseman turns around and runs straight back at a constant velocity V=7.00m/s, for a time t=2.00s. He then catches the ball at the same height at which it left the bat. The third baseman was initially L=18.0m from the location where the ball was hit at home plate.  

1) Find a vector expression for the velocity v  of the softball 0.100 s 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.

2) Find a vector expression for the position r  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 yare 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

the initial velocity of the ball in y direction is

vy = g/t

=9.8/1.0 s

= 9.8 m/s

here time take 1.0 s since  it takes 2 seconds for the ball to go up and then back down,

For the x velocity, the third baseman runs for 2 seconds at a speed of 7 m/s.

d= vt = 7 m/s * 2 s = 14 m

total distance = d+ L = 14 m + 18 m = 32 m

The ball covered this distance in 2 seconds, so it must have been going 16 m/s in the x direction.

v= sqrt vx^2 + vy^2 = sqrt ( 16^2 + 7^2 ) = 18.77 m/s

The only component of velocity will change from start to finish is the y velocity. Since it takes 1 second for the ball to reach its maximum height .

it also takes 1 second for it to fall back down. This means that 0.1 s before the ball is caught, it will have been falling for 0.9 s, and

so it’s speed in the y direction is

vy = (0.9 * 9.807) = -8.8263 m/s.

vx= 16 m/s , vy = -8.82m/s

(2)

Since the ball will have been moving for 1.9 s

the position covered in x direction is

xi = 16 m/s ( 1.9)= 30.4 m

Apply kinematic equation

Vf^2 = Vi^2 + 2a(s)
9.807^2 = 8.8263^2 + 2 * 9.807 * s
96.2 = 77.9 + 19.614 * s

s = 18.3 / 19.614
s = 0.933 m

(x,y) = (30.4, 0.932 m)

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