Question

A 34.9-kg child starting from rest slides down a water slide with a vertical height of 16.0 m. (Neglect friction.)

(a) What is the child's speed halfway down the slide's vertical distance?

(b) What is the child's speed three-fourths of the way down?

Answer 1

(a) 12.52 m/s

(b) 15.34 m/s

Step-by-step explanation:

mass, m = 34.9 kg

h = 16 m

(a) Initial velocity, u = 0

height = h / 2 = 16 / 2 = - 8 m (downward)

let the speed of child is v.

acceleration, a = - 9.8 m/s^2 (downward)

Use third equation of motion

`v^(2)=u^(2)+2as`

`v^(2)=0^(2)+2* 9.8* 8`

v = 12.52 m/s

(b) Initial velocity, u = 0

height = 3 h / 4 = 12 m = - 12 m (downward)

let the speed of child is v.

acceleration, a = - 9.8 m/s^2 (downward)

Use third equation of motion

`v^(2)=u^(2)+2as`

`v^(2)=0^(2)+2* 9.8* 12`

v = 15.34 m/s

Answer 2

a) 12.528 m/s

b) 15.344 m/s

Step-by-step explanation:

Given:

Mass of the child, m = 34.9 kg

Height of the water slide, h = 16.0 m

Now,

a) By the conservation of energy,

loss in potential energy = gain in kinetic energy

mgh =
`\frac{\textup{1}}{\textup{2}}\textup{m}*\textup{v}^2`

where,

g is the acceleration due to the gravity

v is the velocity of the child

thus,

at halfway down, h =
`\frac{\textup{16}}{\textup{2}}`= 8 m

therefore,

34.9 × 9.81 × 8 =
`\frac{\textup{1}}{\textup{2}}\textup{34.9}*\textup{v}^2`

or

v = 12.528 m/s

b)

at three-fourth way down

height =
`\frac{\textup{3}}{\textup{4}}*16` = 12 m

thus,

loss in potential energy = gain in kinetic energy

34.9 × 9.81 × 12 =
`\frac{\textup{1}}{\textup{2}}\textup{34.9}*\textup{v}^2`

or

v = 15.344 m/s