A ball is dropped from a high rise platform at t=0 starting from rest.After 6s another ball is thrown downwards from the same platform with a speed 'v'.The 2 balls meet at t=18s.What is the value of 'v'?

Question

1462 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-08T00:17:20

#v_i=73.5 m/s#

I will be using #d=v_it+1/2at^2#

the first ball is dropped so the initial velocity is #0m/s#, so the equation can be modified to:

#d= 1/2at^2#

For the second ball, we don't know the initial velocity of the ball but we are trying to find it, so we will still use:

#d=v_it+1/2at^2#

Set the formulas equal to each other as the balls have the same distance travelled at #t=18# seconds:

Note that the time value we will use for Ball 2 is 12 seconds, as it was dropped 6 seconds later:

#vit+1/2at^2= 1/2at^2#

#vi(12 s)+1/2(9.8 m/s^2)(12 s)^2= 1/2(9.8 m/s^2)(18 s)^2#

#v_i(12s)= 882m#

#v_i=73.5 m/s#