Question

OKAY theres three parts. First part asks to

a) calculate the distance she travels along the incline before landing (answer in units of m)

b) Determine how long the ski jumper is airborne (answer in units of s)

c) what is the magnitude of the relative angle with which the ski jumper hits the slope? (answer in units of degree)

PLEASE HELP OR IM GONNA CRY TYSM

Answer 1

a)
`d \approx 343.260\,m`, b)
`\Delta t = 7.573\,s`, c)
`\gamma = 15.706^(\textdegree)`

Step-by-step explanation:

a) The skier is experimenting an parabolic motion. The kinematic expressions for the skier are:

`x = v_(x,o) \cdot \Delta t`

`0 = y - (1)/(2) \cdot g \cdot (\Delta t)^(2)`

From trigonometry, there is the following relationship between vertical and horizontal components:

`y = x \cdot \tan \alpha`

The following expression is constructed by substituting
`x` and
`\Delta t` in the vertical motion expression and simplifying the resultant equation:

`0 = x\cdot \tan \alpha - (1)/(2)\cdot g \cdot \left((x)/(v_(x,o)) \right)^(2)`

`x \cdot \left(x-(2\cdot v_(x,o)^(2)\cdot \tan \alpha)/(g) \right) = 0`

The roots of the second-order polynomial are:

`x_(1) = 0\,m` and
`x_(2) = (2\cdot v_(x,o)^(2)\cdot \tan \alpha)/(g)`

The second one represents the horizontal distance travelled by the skier:

`x = (2\cdot v_(x,o)^(2)\cdot \tan \alpha)/(g)`

`x = (2\cdot \left(26\,(m)/(s) \right)^(2)\cdot \tan 55^(\textdegree))/(9.807\,(m)/(s^(2)) )`

`x = 196.886\,m`

The vertical distance travelled by the skier is:

`y = (196.886\,m)\cdot \tan 55^(\textdegree)`

`y = 281.182\,m`

The distance that skier travels along the incline is computed through the Pythagorean Theorem:

`d = \sqrt{(281.182\,m)^(2)+(196.886\,m)^(2)}`

`d \approx 343.260\,m`

b) The time spent by the skier is:

`\Delta t = (x)/(v_(x,o))`

`\Delta t = (196.886\,m)/(26\,(m)/(s) )`

`\Delta t = 7.573\,s`

c) The final vertical velocity of the skier is:

`v_(y) = -\left(9.807\,(m)/(s^(2))\right)\cdot (7.573\,s)`

`v_(y) = -74.268\,(m)/(s)`

The absolute angle of the velocity is:

`\beta = \tan^(-1) \left((74.268\,(m)/(s) )/(26\,(m)/(s) ) \right)`

`\beta \approx 70.706^(\textdegree)`

The magnitude of the relative angle with which the ski jumper hits the slope is:

`\gamma = \beta - \alpha`

`\gamma = 70.706^(\textdegree) - 55^(\textdegree)`

`\gamma = 15.706^(\textdegree)`

Answer 2

(a)343.50 m

(b)10.82 s

(c)21.23

Step-by-step explanation:

The x and y cordinates are
`dcos\theta` and
`dsin\theta` respectively

The horizontal distance travelled,
`x=v_(ox)t=dcos\theta`

Making t the subject,
`t=(dcos\theta)/(v_(ox))`

Since
`y=0.5gt^2=dsin\theta`, we substitute t with the above and obtain

`0.5g((dcos\theta)/(v_(ox)))^2=dsin\theta`

Making d the subject we obtain

`d=(2v_(ox)^2sin\theta)/(gcos^2\theta)`

`d=(2*26^2sin55)/(9.8cos^255)`

d=343.5046

d=343.5m

(b)

Already derived,
`t=(dcos\theta)/(v_(ox))`

hence

`t=(343.50cos55)/(26)}`

t=10.82 s

(c)

`v_(ox)=v_(x)`=26m/s

`v_(y)=v_(oy)-gt`=0-9.8*10.82=-106.04

`\theta_(t)=tan^(-1)((v_(y))/(v_(x)) )=tan^(-1)((-106.04)/(26) )`

`tan^(-1)=(-106.04)/(26)`= -76.23

-76.23+55=-21.23

Therefore angle is
`21.23^(0)`