On a hot summer day, a young girl swings on a rope above the local swimming hole. when she lets go of the rope her initial velocity is 2.25m/s at an angle of 35º above the horizontal. if she is in flight for 1.20s, how high above the water was she when she let go of the rope?

5.52 m

Explanation:

Using the equation;

Vy = Vo sin θ

    Where; Vo = 2.25 m/s and θ = 35°

We get;

       = 2.25 sin 35°

        = 1.29 m/s

Thus; Vy = 1.29 m/s

But a = -9.81 m/s (against gravity)

      t = 1.20 s and

y = Vyt + 1/2 at²

   where y is the height above the water

  = 1.29 × 1.2 + (1/2 × -9.81 × 1.2²)

  = 1.548 + (-7.0632)

   = - 5.5152 m

   Thus; the height of the girl above the water is 5.52 m

its c its is a mixture of gases

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