Question

A ball is dropped from rest from the top of a building with height 50 meters. At the same instant, you throw a ball upward from ground level carefully aimed to hit the falling ball. If the initial speed of your throw is 19 m/s, then the two balls collide at a height above the ground equal to:

a) 4.50 m b) 7.61 m c) 12.2 m d) 16.1 m e) 2.15 m

Answer #1

according to the concept of the motion in one dimensional

s=ut+1/2at^2

Given that

height h=50 meter

initial velocity of the first ball u1=0

initial speed of second ball u2=19 m/s

now we find the height above from the ground they meet

ball1=>50-x=0+1/gt^2.................(1)

ball2=>x=19t-1/2gt^2..................(2)

solve eq1 &eq2

50-1/2gt^2=19t-1/2gt^2

50=19t

time t=2.632 sec

=>50-x=1/2*9.8*2.632^2

50-x=33.94

height x=16.06=16.1 m

the corrrect option D

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