Question

You toss a ball straight up at 6.8 m/s ; it leaves your hand at 2.0 m above the floor. Suppose you had tossed a second ball straight down at 6.8 m/s (from the same place 2.0 m above the floor). When would the second ball hit the floor?

Answer 1

Answer:0.249 s

Step-by-step explanation:

Given

Ball is tossed down with a velocity of 6.8 m/s downward

height from ground=2 m

therefore time to reach ground is

`s=ut+(gt^2)/(2)`

`2=6.8* t+(9.81* t^2)/(2)`

`9.81t^2+13.6t-4=0`

`t=(-13.6\pm √(13.6^2+4* 4* 9.81))/(2* 9.81)`

`t=(-13.6+18.49)/(19.62)=0.249 s`