Max Height for Pole-Vaulter with Speed 6.00 m/s

[iceT]

A pole-vaulter approaches the takeoff point at a speed of 6.00 m/s. Assuming that only this speed determines the height to which he can rise, find the maximum height at which the vaulter can clear the bar.
in this 1 I am lost......
I ONLUY GOT 1 GIVINGS (6.00 m/s )
should i use the ...h0-hf = m ??


person is making homemade ice cream. She exerts a force of magnitude 19 N on the free end of the crank handle, and this end moves in a circular path of radius 0.32 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.9 s, what is the average power being expended?


i know that the P= WORK/TIME...
AND P= CHANGE IN ENERGY/ TIME

I TRIDE BOTH ..BUT NONE OF THEM WORKED FOR ME...

 

[Doc Al]

A pole-vaulter approaches the takeoff point at a speed of 6.00 m/s. Assuming that only this speed determines the height to which he can rise, find the maximum height at which the vaulter can clear the bar.
in this 1 I am lost......
I ONLUY GOT 1 GIVINGS (6.00 m/s )
should i use the ...h0-hf = m ??

Think in terms of energy conservation. The vaulter starts with some kinetic energy due to his speed; assume that that energy gets converted to gravitational potential energy to find the maximum height.

person is making homemade ice cream. She exerts a force of magnitude 19 N on the free end of the crank handle, and this end moves in a circular path of radius 0.32 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.9 s, what is the average power being expended?

For every complete turn of the crank, how much work is done? Once you figure out the work, then compute the power.

 

[kyle8921]

I would approach this problem by using the formula for power, which is P = work/time. In this case, the work being done is the force being applied (19 N) multiplied by the distance the handle moves (which is equal to the circumference of the circular path, 2πr). So, the work done in one revolution would be W = 19 N * 2π * 0.32 m = 38π Nm.

Next, we need to find the time it takes for one revolution, which is given as 1.9 s. Now, we can plug these values into the power formula to find the average power being expended:

P = (38π Nm) / (1.9 s) = 20π Nm/s

So, the average power being expended in turning the handle is approximately 20π Nm/s.