027 10.0 points A child throws a steel ball stralght up. Con- slder the motlon o

Question

027 10.0 points A child throws a steel ball stralght up. Con- slder the motlon of the ball after It has left the of 2) 10.0 points the ball (in terrns of the child's hand but before It touches the ground, the ball passes a point A, and assume that forces exerted by the air are e maximum height. H1s e negligible litions, identily the foreets 030 10,0 points A ball is thrown upward trom the ground The aeceleratten of gravity s upward force tre tDZ dow11ward forer of gravity grts closer to eartli es the ball falls har k o th of the earthi

Explanation / Answer

Given

the initial velocity of the ball b1 thrown upward is u = 46 m/s

the second ball b2 dropped from a building of height 27 m

here if b1 covers a distance of x in time t so that b2 can cover a distance of 27-x m in the same time t

writing the equations of motions for b1,b2

for b1

x = ut-0.5*gt^2

for b2

27-x = 0.5*g*t^2

adding the equations we get

x = 46*t-0.5*9.8*t^2

27-x = 0.5*9.8*t^2

--------------------------------------------

27 = 46*t ==> t = 0.587 s

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so the balls can be at same height at time t = 0.587 s

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