A batter hits the baseball A with an initial velocity of v_0 = 103 ft/sec direct

Question

A batter hits the baseball A with an initial velocity of v_0 = 103 ft/sec directly toward fielder B at an angle of 32 degree to the horizontal; the initial position of the ball is 1.6 ft above ground level. Fielder B requires 0.50 sec to judge where the ball should be caught and begins moving to that position with constant speed. Because of great experience, fielder B chooses his running speed so that he arrives at the "catch position" simultaneously with the baseball. The catch position is the field location at which the ball altitude is 7.6 ft. Determine the velocity of the ball relative to the fielder at the instant the catch is made.

Explanation / Answer

Given that
v0 = 103 ft/sec

Altitude = 7.6 ft

Let t be the flight time

now using

y = yo + v*t - 0.5*g*t^2

7.6 = 1.6 +103*sin32*t - 0.5*32.2*t^2

16.1t^2 - 54.58t + 6 = 0

Solve the quadratic equation and obtain t as

t = 3.2751 or 0.1133 sec

Descarding

t = 0.1133 is less than (0.50 sec)

We have

t = 3.2751 sec

Range

R = x0 + vx0*t

= 0 + 103*cos32*3.2751

= 286.07 ft

So fielder must run a total distance of

286.07 - 224 = 62.07 ft

in a period of (3.2751-0.50=2.77)

vB = (62.54 / 2.77)

= 22.57 ft/s

Velocity components of ball when caught

vx = vx0 = 103*cos32

= 87.34 ft/s

vy = vy0 - g*t

= 103*sin 32 - 32.28*3.27

= -50.97 ft/s

So
vab = va - vb

= (87.34i - 50.97 j) - 22.57i

= 64.77 i - 50.97 j ft/s

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