Five-Part Physics Question - Help and Answers Appreciated

Question

Answer 1

(a)

Assuming that her speed at the top of the slide is 0,

her mechanical energy consists only of her potential energy

mgh

(b)

Let v be her speed at the launch site.

Her mechanical energy at the top, mgh, gets converted at the launch site into

potential energy, mgh/5, and

kinetic energy mv²/2

That is,

mgh = mgh/5 + mv²/2

v = √(8gh/5)

(c)

The speed obtained in (b) can be decomposed into a vector sum

v = v_x + v_y

of horizontal and vertical speeds, where

v_x = cos(θ)v,

v_y = sin(θ)v.

While her vertical speed is affected by gravity, her horizontal speed remains constant.

(Ignoring air resistance.)

At the maximum height point, her vertical speed is 0, and her total speed thus equals

her horizontal speed

v_x = cos(θ)v = cos(θ)√(8gh/5)

(d)

By conservation of energy,

her original potential energy, mgh, gets converted at the maximum height point into

potential energy, mgy_max, and

kinetic energy, mv_x²/2

That is,

mgh = mgy_max + mv_x²/2

y_max = h - v_x²/2g

(e)

Let t be the time it takes her from the launch point to the splash point.

During that time she will cover the distance

v_xt.

Thus we just need to calculate the time t.

And for that we will first calculate the vertical component v_y' of her velocity at the

splash point.

At the splash point, all her initial potential energy, mgh, got converted into her

kinetic energy.

The square of her velocity at the splash point is v_x² + v_y'².

Thus the conservation of energy statement is

mgh = m(v_x² + v_y'²)/2

From that

v_y' = √(2gh - v_x²)

Her upward vertical velocity at launch point, v_y, changed into her downward vertical

velocity, v_y', due the gravitational acceleration. Thus

v_y + v_y' = gt

Hence

t = (v_y + v_y')/g

Finally the distance of the splash point from launch point is

v_xt,

where v_x and t can be substituted from the above equations.