A34.9-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?

a) 12.528 m/s

b) 15.344 m/s

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 =

where,

g is the acceleration due to the gravity

v is the velocity of the child

thus,

at halfway down, h = = 8 m

therefore,

34.9 × 9.81 × 8 =

or

v = 12.528 m/s

b)

at three-fourth way down

height = = 12 m

thus,

loss in potential energy = gain in kinetic energy

34.9 × 9.81 × 12 =

or

v = 15.344 m/s

(a) 12.52 m/s

(b) 15.34 m/s

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 = 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 = 15.34 m/s

it would be kevlar ^^

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.