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.
