Question

A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.905 m, as the drawing shows. At the top of the swing his speed is momentarily zero. Ignoring friction and treating the gymnast as if all of his mass is located at his waist, find his speed at the bottom of the swing.

Answer 1

Answer:

5.959 m/s

Explanation:

m = Mass of gymnast

u = Initial velocity

v = Final velocity

$h_i$ = Initial height

$h_f$ = Final height

From conservation of Energy

$\frac{1}{2}mv^2+mgh_f=\frac{1}{2}mu^2+mgh_i\\\Rightarrow\frac{1}{2}mv^2+mg0=\frac{1}{2}m0^2+mgh_i\\\Rightarrow \frac{1}{2}mv^2=mgh_i\\\Rightarrow v=\sqrt{2gh_i}$

$h_i=2r$

$v=\sqrt{4gr}\\\Rightarrow v=\sqrt{4\times 9.81\times 0.905}\\\Rightarrow v=5.959\ m/s$

Velocity of gymnast at bottom of swing is 5.959 m/s

Answer 2

The speed of the gymnast at the bottom of the swing is 5.96 m/s.

Given the following data:

• Distance, r = 0.905 meter.

Scientific data:

• Acceleration due to gravity (g) = 9.8 $m/s^2$.

To determine the speed of the gymnast at the bottom of the swing, we would apply the law of conservation of energy:

How to calculate the final speed.

In accordance with the law of conservation of energy, the potential energy and kinetic energy possessed by the gymnast at the top is equal to the potential energy and kinetic energy possessed by the gymnast at the bottom of the swing:

$P.E_i +K.E_i=P.E_f +K.E_f\\\\mgh_i+\frac{1}{2} mv^2=mgh_f+\frac{1}{2} mv^2\\\\V=\sqrt{2gh} \\\\V=\sqrt{2g \times 2r}\\\\V=\sqrt{4gr}\\\\V=\sqrt{4\times 9.8 \times 0.905}$

Final speed, v = 5.96 m/s.

Read more on kinetic energy here: https://brainly.com/question/1242059