- #1

Zem

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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

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

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