Question

Question #ba0bf

Answer 1

See the solution below.

Explanation:

Assuming that there is no drag experienced by the ski jumper. Also assume that Cartesian coordinate system has its origin at the point when jumper leaves the ramp.

At the time of jump, the ski jumper has given velocity `v_0`.
It has two components `v_(0x)=10 xxcos 15^@ m//s` and `v_(0y)=10 xxsin 15^@ m//s`. Both the components are orthogonal to each other and therefore independent.

It is clear that initially he will rise up and then start falling down due to action of gravity. The acceleration due to gravity `g=9.8 ms^-2` and acts in the direction given by `- haty`. Let the jumper hit the ramp at a distance `h` from the origin.

(a) Imagine a right angled triangle where `d` is its hypotenuse and `h` is its perpendicular. From this figure it is clear that `h=d.sin 50^@`, (`d` in meters).
Hence `h = 0.766d` `m`

(b) The `y` component of the velocity at the time of landing can be calculated with the help of the following equation.
`v^2-u^2=2gh`

Inserting the `y` component of initial velocity `v_(0y)` as calculated above

`v_y^2-(10xxsin 15^@)^2=2xx(9.8)times 0.766(d)`

Inserting value of sin of the angle from the table
`v_y^2=6.6987+15.0136d`
`v_y=sqrt(6.6987+15.0136d)`

There is no change in the `x` component of skier's velocity just before landing.
`v_(0x)=10 xxcos 15^@ m//s=9.6593m//s`

Answer 2

d=48.051 meters
`v_x=9.659 m/s`
`v_y=-28.52 m/s`

Explanation:

a)
`v_x=10*cos15=9.659 m/s` initial velocity x component
`v_y=10*sin15=2.588 m/s` initial velocity y component
`v_xt=d*cos50` , t:flying time ( `v_x` not changes during event)
`t=d*0.643/9.659`, `t=d0.066 s`
`y=d* sin50 + v_y t-(g t^2)/2`
`y= 0` , when landing on bottom
`0=d 0.766 +2.588 t -(g t^2)/2` , `g=9.81 m/s^2`
`0=d* 0.766 + 2.588 d 0.066-9.81(d 0.066)^2/2`
`d=48.051 meters`

b)
`t=d 0.066=48.051 * 0.066=3.171 seconds`
`v_x=9.659 m/s` (no change)
`v_y=2.588-9*t`
`v_y=2.588-9.81*3.171`
`v_y=2.588-31.108`
`v_y=-28.52 m/s`

Answer 3

(a)

`d=27.36"m"`

(b)

`v_x=9.66"m/s"`

`v_y=20.43"m/s"`

Explanation:

(a)

The key to these projectile problems is to treat the vertical and horizontal components of the motion separately and use the fact that they share the same time of flight.

Fig 1

I will use the vertical and horizontal components to eliminate `t` and find the vertical height `h` as in fig. 1

Then by simple trigonometry I will find `d`.

Considering the vertical component I will use:

`s=ut+1/2at^2`

I will use the convention that "up" is -+ve.

This becomes:

`h=10sin(15)t-1/2"g"t^2" "color(red)((1))`

Considering the horizontal component, for which the velocity is constant, we can write:

`10cos(15)=r/t" "color(red)((2))`

From the geometry of the system we can also write:

`tan(50)=h/r`

`:. r=h/(tan(50))" "color(red)((3))`

Now we can set about eliminating `t` which is common to both vertical and horizontal motion:

From `color(red)((2)):`

`t=r/(10cos(15))=r/9.695`

From `color(red)((3)):`

`r=h/(tan(50))=h/1.1917`

Combining these we get:

`t=h/(1.1917xx9.659)=h/11.51`

We can now substitute this value of `t` into `color(red)((1))` and thus eliminate it so `color(red)((1))` becomes `rArr`

`h=10sin(15).(h)/(11.51)-1/2xx9.8xx((h)/(11.51))^2`

`:.h=(2.588h)/(11.51)-(9.8xxh^2)/(2xx132.48)`

`:.h=0.2248h-0.03698h^2`

`:.0.03698h^2+0.7752h=0`

This is a quadratic equation for which:

`h=(-0.7752+-sqrt(0.6))/(2xx0.03698)`

`h=(-0.7752+-0.7752)/(0.07396)`

Ignoring the zero root:

`h=-1.5504/0.07396=-20.96"m"`

The minus sign signifies the downward direction.

From fig.1 we can say that:

`sin(50)=h/d`

`:.d=h/sin(50)=20.94/0.766`

`d=27.33"m"`

(b)

For the horizontal component of velocity:

`v_x=10cos(15)=9.66"m/s"`

For the vertical component we can use:

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

This becomes:

`v_y^2=u_y^2-2gh`

`v_y^(2)=(10sin(15))^2-2gh`

`v_y^2=10sin(15)^2-2xx9.8xx(-20.96)`

`v_y^2=417.51`

`v_y=20.43"m/s"`