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

conduction, convection, radiation

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s