Question

When the skater starts 7 mm above the ground, how does the speed of the skater at the bottom of the track compare to the speed of the skater at the bottom when the skater starts 4 mm above the ground?

Answer 1

Speed is higher and 1.32 times greater.

Step-by-step explanation:

Considering downward motion as positive.

Given:

Case 1:

Initial height (h₁) = 7 mm = 0.007 m [1 mm =0.001 m]

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u₁) = 0 m/s

Final velocity (v₁) = ?

Using conservation of energy, we have:

Decrease in potential energy = Increase in Kinetic energy

`mgh_1=(1)/(2)mv_1^2\\\\v_1=√(2gh_1)=√(2* 9.8* 0.007)=0.37\ m/s`

Case 2:

Initial height (h₂) = 4 mm = 0.004 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u₂) = 0 m/s

Final velocity (v₂) = ?

Using conservation of energy, we have:

Decrease in potential energy = Increase in Kinetic energy

`mgh_2=(1)/(2)mv_2^2\\\\v_2=√(2gh_2)=√(2* 9.8* 0.004)=0.28\ m/s`

From the above values, we can conclude:

`v_1>v_2`

Also,

`(v_1)/(v_2)=(0.37)/(0.28)=1.32\\\\v_1=1.32v_2`

So, the velocity in the first case is 1.32 times greater than velocity in second case.