Question #ba0bf

3 Answers
Jan 25, 2016

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

Jan 31, 2016

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

Feb 5, 2016

(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

MFDocsMFDocs

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"