Question

Starting from rest at the top, a child slides down the water slide at a swimming pool and enters the water at a final speed of 10.0 m/s. At what final speed would the child enter the water if the water slide were four times as high

Answer 1

The child will enter the water at a final speed of 20 m/s

Step-by-step explanation:

Given;

final speed of the child,
`V_f` = 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

`(1)/(2)m (V_f^2 -V_i^2) = mgh\\\\(1)/(2) (V_f^2 -V_i^2) = gh\\\\(1)/(2) (V_f^2 -0^2) = gh\\\\(1)/(2) V_f^2 = gh\\\\V_f^2 = 2gh\\\\g = (V_f^2)/(2h) \\\\(V_f_1^2)/(2h_1) = (V_f_2^2)/(2h_2) \\\\From \ the \ given \ question, \ h_2 = 4h_1\\\\(V_f_1^2)/(2h_1) = (V_f_2^2)/(2(4h_1)) \\\\(V_f_1^2)/(2h_1) = (V_f_2^2)/(8h_1) \\\\(V_f_1^2)/(2) = (V_f_2^2)/(8)\\\\4V_f_1^2 = V_f_2^2\\\\V_f_2 = √(4V_f_1^2) \\\\V_f_2 = 2*V_f_1\\\\V_f_2 = 2*10 \ m/s\\`

`V_f_2 = 20 \ m/s`

Therefore, the child will enter the water at a final speed of 20 m/s if the water slide were four times as high