# Determine the maximum airborne height of the girl in the following case:A child slides without Friction from a height of 10.3 m above the water along a curved water slide. She is launched from a...

Determine the maximum airborne height of the girl in the following case:

A child slides without Friction from a height of 10.3 m above the water along a curved water slide. She is launched from a height 6.1m above the above the water at an angle of 48 degrees above the horizontal toward a pool.

The acceleration of gravity is 9.81 m/s^2.

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### 2 Answers

A comment on answer 1 by justaguide:

The girl is launched at 48 degrees above horizontal and hence has a velocity in the vertical direction and velocity in the horizontal direction.

*The kinetic energy due to velocity in the vertical direction is converted back to potential energy to attain the maximum airborne height after the launch.*

**The kinetic energy due to velocity in the horizontal direction is retained by the girl during her flight and there is no change in that while the girl is airborne.**

*Hence the maximum airborne height is not 9.3 m above the water after the launch.*

Let X be the mass of child and v the velocity at launching from the slide

acceleration due to gravity = 9.81 m/s^2

Angle of launch = 48 dgrees above the horizontal

Height at start of slide = 10.3 m

Height at end of Slide = 6.1 m

Loss in height = 10.3 - 6.1 = 4.2 m

Loss in potential energy = X*g*h

Gain in kinetic energy = 0.5*X*v^2

0.5*X*v^2 = X*g*h

0.5*v^2 = 9.81*4.2

v^2 = 82.404

v = 9.078 m/s

Vertical component of launching velosity = v*sin(48)

= 9.078*0.6691 = 6.074

We know that v^2 - u^2 = 2*a*s, where v is final velocity = 0 at highest point, u = initial velocity (Launching velocity), a = acceleration (-9.81 due to upward motion a launch) and s = distance (from launching point). Substituting these values:

0^2-6.074^2 = 2*(-9.81)*s

s = -6.074^2/(-2*9.81) = 1.9 m

**Maximum airborne height above water **= 6.1+1.9 **= 8.0 m**