Answer:
The child will enter the water at a final speed of 20 m/s
Explanation:
Given;
final speed of the child, [tex]V_f[/tex] = 10 m/s
let the height of water slide, = h
According to conservation of energy, the change in kinetic energy of the child must be equal to the change in potential energy,
ΔK.E = ΔU
[tex]\frac{1}{2}m (V_f^2 -V_i^2) = mgh\\\\\frac{1}{2} (V_f^2 -V_i^2) = gh\\\\\frac{1}{2} (V_f^2 -0^2) = gh\\\\\frac{1}{2} V_f^2 = gh\\\\V_f^2 = 2gh\\\\g = \frac{V_f^2}{2h} \\\\\frac{V_f_1^2}{2h_1} = \frac{V_f_2^2}{2h_2} \\\\From \ the \ given \ question, \ h_2 = 4h_1\\\\\frac{V_f_1^2}{2h_1} = \frac{V_f_2^2}{2(4h_1)} \\\\\frac{V_f_1^2}{2h_1} = \frac{V_f_2^2}{8h_1} \\\\\frac{V_f_1^2}{2} = \frac{V_f_2^2}{8}\\\\4V_f_1^2 = V_f_2^2\\\\V_f_2 = \sqrt{4V_f_1^2} \\\\V_f_2 = 2*V_f_1\\\\V_f_2 = 2*10 \ m/s\\[/tex]
[tex]V_f_2 = 20 \ m/s[/tex]
Therefore, the child will enter the water at a final speed of 20 m/s if the water slide were four times as high
