Sam fields a baseball hit to him in the left field. He then throws the ball to t
ID: 2116158 • Letter: S
Question
Sam fields a baseball hit to him in the left field. He then throws the ball to third to force out the base runner, Mike. Sam releases the ball 1.80 m above the ground with a vertical velocity of 8 m/s and a horizontal velocity of 25 m/s. At the instant Sam releases the ball, he is 41 m from the third baseman, Charlie, and Mike is 13 m from third base and running at 8 m/s toward third. Assume that air resistance does not affect the flight of the ball when answering the following questions.
a. How high in the air does the ball go?
answer:5.06 m
I need workout steps please.
b. How much time does it take for the ball to reach the third baseman?
answer: 1.64 s
I need workout steps please.
c. How high is the ball when it reaches the third baseman?
answer: 1.73 m
I need workout steps please.
d. If Mike maintains a constant velocity of 8 m/s toward third base, does he reach third base before the ball reaches the third baseman?
answer:yes
I need and explanation please.
Explanation / Answer
a) The maximum height of the ball occurs when the y-velocity is zero. vy=v0y+a*tmax. tmax=v0y/a= -v0y/g.
The height of the ball at tmax is y=0.5*a*tmax^2+v0y*tmax+y0=-0.5*v0y^2/g+v0y^2/g+y0=0.5*v0y^2/g+y0=0.5*64/9.8+1.8 = 5.06m,
b)
t is the time it takes the ball to reach the third baseman. That is, distance/rate. 41m/(25m/s)=1.64s.
c) The height of the ball at that time is is y=0.5*a*t^2+v0y*t+y0=0.5*(-9.8)*1.64^2+8*1.64+1.8=1.74m, with difference in hundredths being due to rounding.
d) Mike arrives in 13m/8m/s = 1.625s. 1.625 < 1.64 so he makes it to third safe.
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