The first picture is the example that I am trying to solve. The answer is xf= 89

Question

The first picture is the example that I am trying to solve.  The answer is xf= 89.3 and yf= -62.5 which is not what I got!  I have no idea what I did wrong!  Please look at my solution and tell me how I can fix my answer.  Any help is greatly appreciated!  Thank you!

A ski jumper leaves the ski track moving in the horizontal direction with a speed of 25.0 m/s as shown in Figure 3.12. The landing incline below her falls off with a slope of 35.0degree. Where does she land on the incline? Conceptualize We can conceptualize this problem based on memories of observing winter Olympic ski competitions. We estimate the skier to be air borne for perhaps 4 s and to travel a distance of about 100 m horizontally. We should expect the value of d, the distance traveled along the incline, to be of the same order of magnitude. Categorize We categorize the problem as one of a particle in projectile Figure 3.12 (Example 3.3) A ski jumper leaves the track moving in a horizontal direction. Vx = Vcostheta = 25cos(35)degree 20.48 Vy = Vsintheta = 25sin(35)degree 14.34 x1 = ? x1 = 0 v12 = v12 + 2a(x2-x1) phi = (20.48)2 + 2(9.8)(xp) = 4.9.4 = 19.6xf 4.9.4 = 19.6xf 21.4m = xf Vp2 = v12 + 20(yf-y1) vp = (14.34)2+2(9.8)(xp) 0 = 205.4 + 19.4yf yf = -10.5

Explanation / Answer

Vox = 25 m/s
voy = 0

y = 0.5*g*t^2 --(1)

x = vox*t

==> t = x/vox

substitute t in eqn (1)
we get

y = 0.5*g*(x/vox)^2

y = 0.5*g*(1.43*y)^2/vox^2

y = 0.5*9.8*2.04*y^2/25^2

==> y = 625/(0.5*9.8*2.04) = 62.52 m since it is downward we take -62.52 m

so x = y*1.43 = 89.4 m

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