Free fall velocity and distance

[haruspex]

Vi = 4m/s I think it was all the time vertical velocity

Yes, it was 4m/s, but it only had that vertical velocity at one moment, the moment when it was thrown.
So the 1.9s was from the throw to landing on the ground. You are asked for the time from the top of the arc to landing on the ground.

 

[poh]

Yes, it was 4m/s, but it only had that vertical velocity at one moment, the moment when it was thrown.
So the 1.9s was from the throw to landing on the ground. You are asked for the time from the top of the arc to landing on the ground.

it's take same amount of time to go up as go to down. -1.1s is from ground level so when 1.9s-1.1s its where they meet and point is 0 and the diffrence is time from the top of the arc to ground,

 

[haruspex]

it's take same amount of time to go up as go to down. -1.1s is from ground level so when 1.9s-1.1s its where they meet and point is 0 and the diffrence is time from the top of the arc to ground,

I have read that several times and still cannot decipher it.
There are three points in time:

1. Nut is thrown; height 10m, vertical speed +4m/s
2. Nut reaches highest point; vertical speed 0
3. Nut hits ground; height 0

You found that the time from 1 to 3 is 1.9s, and the time from 1 to 2 is 0.41s. In b), you are asked for the time from 2 to 3.

 

[poh]

I have read that several times and still cannot decipher it.
There are three points in time:

1. Nut is thrown; height 10m, vertical speed +4m/s
2. Nut reaches highest point; vertical speed 0
3. Nut hits ground; height 0

You found that the time from 1 to 3 is 1.9s, and the time from 1 to 2 is 0.41s. In b), you are asked for the time from 2 to 3.

I see now Thank you :D

 

[neilparker62]

Alternate method using average velocity with upwards positive so g = -9.8 ms^-2 and ground level (h) = -10m from projectile launch position. Doesn't add much to what's been discussed above so perhaps more of a summary.

Initial vertical velocity: $$v_i=4\;ms^{-1}$$
Final vertical velocity: $$v_f=-\sqrt{{v_i}^2+2gh}\; ms^{-1}$$
Average vertical velocity: $$ v_{av}=\frac{v_i+v_f}{2}\; ms^{-1}$$
Time of travel: $$ t = \frac{h} {v_{av}} \;s $$
Horizontal displacement: $$ 3t \;m$$

Hence:

a) -4/g
b) t - (-4/g)
c) 3t