A 46 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1 m/s. The acceleration of gravity is...

We use the law conservation of energy to find the solution.

The law of conservation of energy says that the sum of the potential energy at any point of the motion is same. Therefore, if m is the mass of the body (pole vaulter here) , his potential energy at ground level is zero and at a height, h above the ground  the PE = mgh.

His ground velocity is x and the velocity at the height, h  is y, the  kinetic energy at gound is (1/2)mx^2 and at the  height  y is (1/2)my^2. By conservation of energy:

PE+KE of the vaulter at ground =  0 + (1/2)mx^2  = constant k

PE+KE at of the vaulter at height h, = mgh + (1/2)my^2 = constant k

Therefore, from the above two equations and by the lae of conservation of energy:

(1/2)mx^2 = mgh+(1/2)my^2 or

x^2 = 2gh + y^2 , given the ground speed , x=11m/s , speed ar the height h, is y =  1m, we get:

11^2= 2*9.81*h + 1^2. So,

h = (11^2 - 1)/(2*9.81) =  6.1162 meter is the bar height.