A child of mass m starts from rest and slides without friction from a height h a

Question

A child of mass m starts from rest and slides without friction from a height h along a slide next to a pool (see figure). She is launched from a height h/5 into the air over the pool. We wish to find the maximum height she reaches above the water in her projectile motion.

(a) Is the child–Earth system isolated or nonisolated?

(b) Is there a nonconservative force acting within the system?

(c) Define the configuration of the system when the child is at the water level as having zero gravitational potential energy. Express the total energy of the system when the child is at the top of the waterslide. (Use any variable or symbol stated above along with the following as necessary: g.)

(d) Express the total energy of the system when the child is at the launching point. (Use any variable or symbol stated above along with the following as necessary: g and vi for her speed at the launch point.)

(e) Express the total energy of the system when the child is at the highest point in her projectile motion. (Use any variable or symbol stated above along with the following as necessary: g, vx for her speed at maximum height, and ym for her maximum airborne height.)

(f) From parts (c) and (d), determine her initial speed vi at the launch point in terms of g and h.

(g) From parts (d), (e), and (f), determine her maximum airborne height ymax in terms of h and the launch angle . (Use any variable or symbol stated above as necessary.)

(h) Would your answers be the same if the waterslide were not frictionless?

Explanation / Answer

a) Isolated, since there is only her weight which can do work on the child.

b) No, since there is no friction.

c) At the top of the waterslide,
potential energy = mgh & kinetic energy is zero

d)
At the launching point,
potential energy = mgh/5
kinetic energy = 1/2mvi^2

e)
At the point where she lands in the pool.
potential energy = mg*ymax
kinetic energy = 1/2*m*vxi^2

f)
mg (4h/5) = (1/2) m vo²

vo = [8gh/5]
= sqrt[1.6gh]

g)
(1/2) m vox² = mg(h - ymax)

(1/2) vox² = g(h - ymax)

hmax = h (1- 4/5cos^2(theta)

h)
Since there is friction the y max will not be the same.

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