Achild is sitting on the seat of a swing with ropes 10 m long. her father pulls the swing back until the ropes make a 47º angle with the vertical and then releases the swing. if air resistance is neglected, what is the speed of the child at the bottom of the arc of the swing when the ropes are vertical?

7.9 m/s

Explanation:

A right triangle is formed where one leg (let's call it x) form an angle of 47° with the hypotenuse and the hypotenuse is the rope. From cosine definition:

cos 47° = x/10

x = 10*cos 47°

x = 6.82 m

The difference between the rope length and this value gives us the height of the child before her father releases the swing. Let's call this height h.

h = 10 m - 6.82 m = 3.18 m

Before the swing starts the child is not moving, then all of her energy is potential energy (PE). When she gets at the bottom of the arc of the swing,  all of her PE is transformed into kinetic energy (KE).

KE = PE

(1/2)*m*v² = m*g*h

where v is her speed and g is the standard gravitational acceleration. Solving for v:

v = √(2*g*h)

v = √(2*9.81*3.18)

v = 7.9 m/s

if the thermos has a vacuum innner, then heat through conv currents and conduction are reduced = vacuum flask