A girl slides along a curved water slide, without friction, from a height $h$ . What is the maximum airborne height $y$ in terms of $h$ and $θ$ ?

Question

A girl slides along a curved water slide, without friction, from a height h . What is the maximum airborne height y in terms of h and θ ?

TestHub · Accepted Answer

Part 1: Energy Conservation

Initial state (at height $h$ ):

Total Energy ( $E_{1}$ ) = $m g h$

Final state (at height $h /5$ , just before leaving the slide):

Total Energy ( $E_{2}$ ) = $m g (h /5) + \frac{1}{2} m v^{2}$

By conservation of energy: $E_{1} = E_{2}$

$m g h = m g (h /5) + \frac{1}{2} m v^{2}$

$v = \frac{8 g h}{5}$

This is the speed of the girl as she leaves the slide at an angle $θ$ above the horizontal.

Part 2: Projectile Motion

The velocity $v$ has horizontal and vertical components:

Horizontal component: $v_{x} = v cos θ$

Vertical component: $v_{y} = v sin θ$

The maximum additional height ( $y_{a dd}$ ) reached by a projectile is given by the formula:

$y_{a dd} = \frac{v _{y}^{2}}{2 g}$

Substitute $v_{y} = v sin θ$ :

$y_{a dd} = \frac{( v sin θ ) ^{2}}{2 g}$

$y_{a dd} = \frac{v ^{2} sin ^{2} θ}{2 g}$

Now substitute the expression for $v^{2}$ from Part 1:

$y_{a dd} = \frac{( \frac{8 g h}{5} ) sin ^{2} θ}{2 g}$

$y_{a dd} = \frac{8 g h sin ^{2} θ}{10 g}$

$y_{a dd} = \frac{4 h sin ^{2} θ}{5}$

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