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:

Speed is higher and 1.32 times greater.

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

Answer:Speed is higher and 1.32 times greater.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[tex]mgh_1=\frac{1}{2}mv_1^2\\\\v_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 0.007}=0.37\ m/s[/tex]

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[tex]mgh_2=\frac{1}{2}mv_2^2\\\\v_2=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.004}=0.28\ m/s[/tex]

From the above values, we can conclude:[tex]v_1>v_2[/tex]

Also,

[tex]\frac{v_1}{v_2}=\frac{0.37}{0.28}=1.32\\\\v_1=1.32v_2[/tex]

So,

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