Question #ffc23

1 Answer
May 29, 2016

Answer:

#5/4s and 7.8125m#

Explanation:

Let the balls meet t sec after the second ball is thrown.
So the 1st ball will then complete (t+1) sec of its journey.

The imitial velocity of the 1st ball is zero .So it will fall a height #h=1/2*g*(t+1)^2#

The 2nd ball's initial velocity being #30m/s# it should fall the same height during t sec
So #h=30*t+1/2*g*t^2#

Inserting #g=10ms^-2# and equating two height we can write
#1/2*10*(t+1)^2=30t+1/2*10*t^2#
#=>5(t+1)^2-5t^2=30t#
#=>5(2t+1)=30t#
#=>2t+1=6t#
#=>t=1/4s#

So the ball will meet #(t+1)=1/4+1=5/4s# after the 1st ball starts.

During this time it will fall
#h=1/2*10*(5/4)^2=125/16m=7.8125m#