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 4.64 m/s. at what final speed would the child enter the water if the water slide were twice as high? ignore friction and resistance from the air and the water lubricating the slide.

Answer 1

Energy conservation law states:
$E_{total} = E_{potential} + E_{kinetic} $
At the top potential energy is at maximum and kinetic energy is zero. At the bottom potential energy is zero and kinetic energ is at maximum.

We have:
$E_{total-top} = E_{total-bottom} \\ E_{potential-top} + E_{kinetic-top} =E_{potential-bottom} + E_{kinetic-bottom} \\ E_{potential-top} =E_{kinetic-bottom} \\ m*g*h = \frac{1}{2} *m* v^{2} \\ g*h = \frac{1}{2} * v^{2}$

We can use this formula to calculate height:
$h= \frac{\frac{1}{2} * v^{2} }{g} \\ h= \frac{\frac{1}{2} * 4.64^{2} }{9.81} \\ h=1.09m$

To calculate requested speed we just need to double the height and insert it into equation:
$v= \sqrt{2*g*2*h} \\ v=\sqrt{2*9.81*2*1.09} \\ v=6.54 m/s$

Answer 2

Answer:

The Final speed would be 6.5 m/s, if the slide is of twice the initial height.

Explanation:

Given Information

Initial speed $u=0$

Final speed $v= 4.64 \mathrm{ m/s}$

Ignoring the friction and resistance from air and water lubricating the slide.

Consider the conversion of energy.

Loss in potential energy = Gain in kinetic energy

$mgh_{1} -0=\frac{1}{2} mv^{2} -0\\mgh_{1} =\frac{1}{2} mv^{2} \\h_{1}=\frac{v^2}{2g}$

Substituting the value.

$h_{1}=\frac{(4.64)^2}{2\times9.81}\\& = 1.09 \, \mathrm{ m}$

If the height is doubled $h_{2} =2h_{1} =2\times 1.09\, \mathrm { m} = 2.18\, \mathrm { m}$

From the above relation.

$h_{2}=\frac{v_{2}}{2g}\\v_{2}=\sqrt{2gh_{2}$

Substituting the value.

$v_2=\sqrt{2\times9.81\times2.18} \\=6.54 \mathrm {m/s}$

Therefore, the final speed would be $6.54 \mathrm {m/s}$.

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