Question

Ask Your Teacher A 51-kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.

Answer 1

Answer:

6.11 N

Explanation:

Let,

m= mass of vaulter

Vi initial speed =11 m/s

Vf speed on pole (final) =1.1 m/s

Yi = initial height (when on ground)=o m

Yf = Height while crossing the pole =?

This problem can be solved by using conservation of energy rule;

KEi + PEi = KEf + PEf

==> 1/2 m$Vi^{2}$+ mg $Yi_{}$ = 1/2 m $Vf^{2}$+mg$Yf_{}$

Initially the altitude of vaulter is zero i-e $Yi=0_{}$ ==> PEi = mg $Yi_{}$= mg×0=0

==>1/2 m$Vi^{2}$= /2 m $Vf^{2}$+mg$Yf_{}$

==> m$Vi^{2}$=m $Vf^{2}$+2mg$Yf_{}$

cancelling m

==> $Vi^{2}$= $Vf^{2}$+2g$Yf_{}$

==> $Vi^{2}$- $Vf^{2}$=2g$Yf_{}$

==> $Yf_{}$=$\frac{Vi^{2}-Vf^{2} }{2g}$

==> = $=\frac{11^{2 }-1.1^{2} }{2 (9.8)}$

==>  $Yf_{}$=$=\frac{119.79}{19.6}$=6.11N