# 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.

najm1947 | Student

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.

najm1947 | Student

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