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.

Answer 1

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.

Answer 2

a)height of baseball = Uy^2/2g=64/19.6=3.265m (Uy-vertical velociyty=8)

maximum height from ground=1.8+3.265=5.0265

b)time = distance /speed =41/25=1.64 (horizontal speed-25)

c)y=ut-0.5gt^2 (u-vertical velocity(Uy)=8);(time =1.64s ,time taken to reach 3rd baseeman)

y=8*1.64-0.5*9.8*1.64^2=(-0.059) (y-height)

height from ground =1.8+(-0.059)=1.74

d)time =distance /speed=13/8=1.625s

since 1.625<1.64 he reaches before the ball