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 curved waterslide (see figure). She is launched from a height h/5 into the pool.

(d) Determine her initial speed v0 at the launch point in terms of g and h.

(e) Determine her maximum airborne height ymax in terms of h, g, and the horizontal speed at the height, vx.

(f) Use the x-component of the answer to part (d) to eliminate vx from the answer to part (e), giving the height ymax in terms of g, h, and the launch angle .

Explanation / Answer

Take gravitational potential energy(GPE) at the water level of the pool as zero.

Let
M = mass of the child
U1 = GPE of the child when she begins sliding,
U2 = GPE of the child when she is at height h/5 above the pool,
U3 = GPE of the child at max height in air,
v1 = speed of the child when she begins sliding,
v2 = speed of the child when she is at height h/5 above the pool,
v3 = speed of the child at max height in air,
K1 = kinetic energy(KE) of the child when she begins sliding,
K2 = KE of the child when she is at height h/5 above the pool,
K3 = KE of the child at max height in air,

Then
U1 = Mgh--------------------------(1)
v1 = 0
K1 = 0---------------------------------(2)

U2 = Mgh/5--------------------------(3)
K2 = 1/2 * M*v2^2--------------(4)

U3 = Mgy-----------------------------(5)
K3 = 1/2 *M*v3^2--------------(6)

By conservation of energy,
U1 + K1 = U2 + K2
Therefore, from (1),(2),(3),(4):-
Mgh + 0 = Mgh/5 + 1/2 *M*v2^2
Dividing by M,
gh = gh/5 + 1/2 *v2^2
Or gh - gh/5 = 1/2 *v2^2
Or 4gh/5 = 1/2 *v2^2
Multiplying by 2,
8gh/5 = v2^2  <-----------------------------------------------ans d
Or v2 = sqrt(8gh/5)
Horizontal component of v2 = v2*cos(theta)
= sqrt(8gh/5) * cos(theta)
When, flying in air, horizontal component of velocity remains constant. At max height vertical component velocity = 0
Therefore v3 = sqrt(8gh/5) * cos(theta)<------------------------ans e
K3 = 1/2 *M*v3^2
Or K3 = 1/2 *M* [sqrt(8gh/5) * cos(theta)]^2
Or K3 = 1/2 *M* (8gh/5) * cos^2(theta)
Or K3 = M* (4gh/5) * cos^2(theta)

By conservation of energy,
U1 + K1 = U3 + K3
Therefore, from (1), (2), (5), (7): -
Mgh + 0 = Mgy + M* (4gh/5) * cos^2(theta)
Dividing by Mg,
h = y + (4h/5) * cos^2(theta)
Or y = h - (4h/5) * cos^2(theta)
Or y = h[1 - (4/5)cos^2(theta)] <---------------------------max height ans f

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