A book is thrown upward from a height of 10.0 m and lands with a velocity of -17.50 m/s. What was its initial velocity?

Data:

Distance=10m

V(i)=-17.50m/s

V(f)=?

Solution:

we will use third equation of motion to find vf

2aS=Vf^2-Vi^22

2*10*10=Vf^2+17.50^2

200=Vf^2+17.50^2

Vf^2=306.25-200

Vf^2=106.25

taking cube root on both sides we get

Vf=10.30m/s

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min