2-Dimensional motion problem once again

[r3dxP]

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0 degrees below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. the car rolls from rest down the incline with a constant acceleration of 4.0m/s^2 for a distance of 50.0m to the edge of the cliff. the cliff is 30.0m above the ocean.

a. what is the car's position relative to the base of the cliff when the car lands?
b. how long is the car in the air?

this is one of my question for school, and i have no idea what to do.. my attempt was : R=1/2at^2 ; 50.0m=1/2(4.00m/s^2)(t^2); t=5.00s, thus it was on the cliff for 5seconds. now what? i have no clue.. :( any help will be nice..

 

[StNowhere]

If you have the time it took to go down the cliff, then you should be able to get it's velocity once it leaves the cliff. Now it's gravity's turn. Note, of course, the angle with which the car leaves the cliff, which means it has initial velocity in both x and y.

 

[r3dxP]

R=1/2at^2 ; 50.0m=1/2(4.00m/s^2)(t^2); t=5.00s, thus the car was on the cliff for 5seconds. ok, so.. Vi = 50.0m/5.00s = 10.0m/s. this is the initial velocity once the car is off the cliff. In respect to the vertical displacement, x=Vix*t - (1/2)(g)(t^2) ; 30.0m=(10.0m/s)(sin 66.0degrees)(t)-(4.9m/s^2)(t^2); by rearranging, we get (4.9m/s^2)t^2 - (9.14m/s)t + 30.0m = 0; by using the quadratic equation, we get t= 1.92, -.922; thus, t=1.92 because negative value for time does not exist. The car was in the air for 1.92 seconds.

 

[r3dxP]

i just solved for b, is that correct?

 

[StNowhere]

R=1/2at^2 ; 50.0m=1/2(4.00m/s^2)(t^2); t=5.00s, thus the car was on the cliff for 5seconds. ok, so.. Vi = 50.0m/5.00s = 10.0m/s. this is the initial velocity once the car is off the cliff. In respect to the vertical displacement, x=Vix*t - (1/2)(g)(t^2) ; 30.0m=(10.0m/s)(sin 66.0degrees)(t)-(4.9m/s^2)(t^2); by rearranging, we get (4.9m/s^2)t^2 - (9.14m/s)t + 30.0m = 0; by using the quadratic equation, we get t= 1.92, -.922; thus, t=1.92 because negative value for time does not exist. The car was in the air for 1.92 seconds.

If the car was on the cliff for 5 s, you have the acceleration, 4 m/s^2. Thus,
$$v_0 = gt = 4 \ m/s^2 \ * \ 5 \ s = 20 \ m/s$$

The car now falls under gravity from a height of 30 m.
$$30 = 20 \sin{24} \ * \ t + \frac{1}{2} gt^2$$.
That should give you the hangtime.
Horizontal motion is separated from vertical motion. Its initial velocity is 20 cos(24) m/s, but it takes t sec. to travel its horizontal distance.

 

[r3dxP]

what do i do for part A?

 

[StNowhere]

is g = 9.8m/s^2 or -9.8m/s^2?

As usual, it depends on which direction you're looking. Since the (vertical) velocity is directed down, the acceleration is down, and the displacement is down, I chose "down" to be the positive direction.

So use the positive value for g.

 

[StNowhere]

what do i do for part A?

The time the car spends in the air is also the only time the car has to travel horizontally. You have the initial velocity and the angle, so you should be able to calculate the initial horizontal velocity. Once you have that, velocity times time equals ...?

 

[r3dxP]

ah i get 32.6m. but wouldn't i have to add 50sin66 to it because you have to add the distance the car traveled while declining from the cliff to the edge of the cliff?

 

[StNowhere]

ah i get 32.6m. but wouldn't i have to add 50sin66 to it because you have to add the distance the car traveled while declining from the cliff to the edge of the cliff?

That depends on the shape of the cliff as defined in the original problem. I imagined it as a straight drop-off at the end, in which case the base of the cliff is directly below the point where the car leaves the cliff, in which case, no, you don't need to add on, since the question asks for the distance from the base. If the cliff is differently shaped, that would play a role.