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Three kinematics problems

  • Thread starter Zem
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Zem

33
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Air resistance is ignored in all three problems.
1) A particle is launched at a 45-degree angle 1000m from the edge of a clff that is 500m high. It needs to land 1000m from the base of the cliff. What initial velocity is necessary?

Y_f = y final = -500m
x_f = x final = 2000m
g = 9.81m/s^2

I found this equation used for a similar problem, and was wondering if it works in this situation.
y_f = (tan(theta)) * x_f - [g/2 * (v_i^2) * (cos^2(theta)] * x_f^2

I got 125m/s. Is that right? Not sure if I'm using the right formula for this situation.

2) A ski jump is designed so that the skier leaves the jump moving horizontally and then lands h = 10.0m below and D = 20.0m beyond the edge of the jump. Find (a) The time t the skier will remain in the air. (b) The speed v_i required to travel a distance D horizontally. (c) The vertical component of the skier's velocity upon landing. (d) The angle theta of the landing ramp so that the skier lands parallel to the ramp.
v_yi = initial velocity in y direction
v_yf = final y-velocity
theta_landing_ramp = angle of the landing surface
(a) Can I use t = sq_rt(2h/g) since v_yi = 0? I got 1.43s, which seems like a quick jump considering the skier went 20m.
(b)v_xi = delta-x/delta-t = 14m/s?
(c)v_yf = h/t = 7.00m/s ?
(d)theta_landing_ramp = tan^-1(10m/20m) = 26.5 degrees?

3) A ski-jumper leaves the ski track moving the horizontal direction with a speed of 25.0m/s. The landing incline below him falls off with a slope fo 35.0 degrees. Where does he land on the incline?
v_yi = 25.0m/s
x_f = horizontal distance from the edge of the track
y_f = vertical distance
a_y = -g
d = hypotonuese
(1)x_f = v_xi(t) = 25.0m/s * (t)
(2)y_f = v_yi(t) + 1/2(a_y(t)^2) = -4.9m/s^2 * (t)^2

(3) x_f = d * (cos(35.0))
(4)y_f = -d * (sin(35.0))
d * (cos(35.0)) = (25.0m/s)t
-d * (sin(35.0)) = -4.9m/s^2(t)^2

The book says: Solving (3) for t and substituting the result into (4), we find that d = 109m. Then they calculate x_f and y_f:
x_f = d * (cos(35.0))
Y_f = -d * (sin(35.0))

But how did they solve for t and find d = 109m? (3) seems to have two unknowns in only one equation.

All the Best,
Zem
 
Last edited:
208
0
you can use 'R = (v0^2 / g) * sin(2*theta)

R = 2000
g = 9.8
then do the alg

I got v = 140
 

Zem

33
0
Originally Posted by mathmike
you can use 'R = (v0^2 / g) * sin(2*theta)

R = 2000
g = 9.8
then do the alg

I got v = 140
I was thinking the change in height had to be included.
y_f = change in height = -500m
-500m = (tan(45)) * 2000m - [9.8m/s^2/(2v_i^2) * (cos^2(45))] * 4x10^6m
0m = (1) * 2500m - [9.8m/s^2/((2v_i^2) * (cos^2(45))] * 4x10^6m
2500m = 4x10^6m * (19.6/2 * v_i^2)
1600m * (19.6m/s^2) = 2 * v_i^2
15,680m^2/s^2 = v_i^2
v_i = 125m/s

I saw that range formula in another thread, but thought it wouldn't work for this problem because there is a change in height. Am I right?

Thanks!
 
208
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You are using the equation of a projectile for this one. If you use the equation R = (V0^2 / G) *sin(2 theta) you are taking into account the disp in Y.
 
208
0
The answers to question number appers to be correct
 
208
0
The way i did the first one was:

the x_ vel. is v0*cos{@} 1

the y_ vel is v0*sin{@} 2

you can get the x_disp from v0*cos{@}*t 3

and the Y_ disp from v0*sin{@}*t - 1/2 * g* t^2 4

the time is derived from 3 x_disp / v0*cos{@} 5

now you can use the relationship y_disp = x_disp * tan{&} 6

now you can sub 5 and 6 into 4 and you get something like this after simplification

tan{&} = tan {@} - 1/2 * (( g * x_disp ) / (v0^2 * cos{@} ^2)) 7

and then

x_disp = 2 * (v0 * cos{@}) ^2 * (tan{@} - tan{&}) / g) 8

then you use the given values

x_disp = 2000

@ = 0

& = 45

g = -9.8

and you come up with v0 = 98.99 m/s

apparently my first post was in error my greatest apologies

your solution for prob 2 seems to be okay

as for three you can find the answer by the same method as above

i get x_disp = 89.31 m

then use r = x_disp / cos{&} = 109.03 m

you can then plug this back into the eq. for x_disp or y_disp to get t = 3.57 s
 

Zem

33
0
Thanks for your help, Mathmike!

I asked about #1 in the help session with the professor, and it turns out my solution was right. :cool: Although he did say that equation only works when the trajectory starts from the origin, so in keeping with good form, he suggested doing the problem from scratch:

h = 500m
Yf = 0
Xf = 2000m

X = 0 Vi(cos(theta))(t)
t = x/V(sin(theta))
Yf = h + Vi(sin(theta))(t) - 1/2(gt^2)

then:
Yf = h + Vi(sin(theta))(x/Vi(cos(theta))) - 1/2(g(Xf^2/V^2(cos^2(theta))))

Yf - h - Xf(tan(theta)) = -1/2(g(Xf^2/Vi^2(cos^2(theta))))

(h + Xf(tan(theta) - Yf))(2(Vi^2(cos^2(theta)))) = gXf^2

Vi^2 = (gXf^2/2(cos^2(theta))(h + Xf(tan(theta)) - Yf)

Vi = (Xf/(cos(theta)))(sq_rt(g/(2(h + Xf(tan(theta))) - Yf)))
= 2829(sq_rt(0.00196))
= 125m/s
 

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