# Three kinematics problems

#### Zem

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:
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#### mathmike

you can use 'R = (v0^2 / g) * sin(2*theta)

R = 2000
g = 9.8
then do the alg

I got v = 140

#### Zem

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!

#### mathmike

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.

#### mathmike

The answers to question number appers to be correct

#### mathmike

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

I asked about #1 in the help session with the professor, and it turns out my solution was right. 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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