You are standing on top of a building 135 meters tall You th

You are standing on top of a building 135 meters tall. You throwa ball A upward with a velocity of 22.0 m/s. At the exact samemoment a friend throws ball B upward from the ground with velocityof 46.0 m/s. These two balls then collide at some later time.

a) How long after these two balls are released will they collide?(answer: 5.63 s)

b) Where will these two balls be when they collide? (answer: fromground=104 m)

c) what will be the velocity of each ball just as theycollide?(answer: Ball a: -33.1 m/s, Ball b: -9.1m/s)

d) What will be the relative velocity between these balls at themoment they collide? (answer: 24m/s)

WHAT ARE THE STEPS TO DOING THESE PROBLEMS? you shouldget the answers stated after each question.

height of building H = 135 m

After time t both balls are hit And height of the balls whenthey hitted be h

Ball A :

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Initial velocity u = 22 m / s

distance S = H - h = 135 - h

from the relation S = ut - ( 1/ 2) g t ^ 2

-135+h = 22 t - (1/ 2) g t ^ 2

-135+h= 22t - 4.9 t^ 2 -----( 1)

ball B :

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Initial velocity u \' = 46 m / s

distance S \' = h

fom the relation S \' = u\'t -( 1/ 2) gt ^ 2

h= 46 t - 4.9 t^ 2

h=46t -22t -135+h since from eq (1)

= 24 t - 135 + h

0= 24t - 135

t = 5.625 s

plug this in eq ( 1 ) we get -135+h = 22(5.625)-4.9(5.625)^2

h= 135 +(22*5.625) -4.9(5.625)^2

= 103.71 m

You are standing on top of a building 135 meters tall You th

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