What are the Kinematics of a Car Plunging off the Grand Canyon Edge?

AI Thread Summary
The discussion revolves around calculating the kinematics of a car falling off the Grand Canyon, specifically addressing the time to impact and the vertical impact velocity. The car's initial vertical velocity is zero, and the acceleration due to gravity is -9.81 m/s². Participants clarify that the depth of the canyon is 500 m, and the first equation of motion can be used to solve for the time of fall. There is confusion regarding the final velocity being negative, but it is emphasized that the car is in free fall with no initial vertical velocity. The conversation concludes with a focus on applying the correct equations to find the time and final velocity accurately.
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Homework Statement



Thelma and Louise's car falls off the edge of the Grand Canyon and plunges to certain catastrophe below. The depth of the canyon at this point is 500 m. (Assume it drops off)

A) How long will it take the vehicle to impact the canyone bottom?

B) What will the vertical impact velocity be? Do you think they will survive, even with air bags?

Homework Equations



d = vit + 1/2at^2

d = (vi+vf)t
2

The Attempt at a Solution



d - 500m
ag = -9.81 m/s^2
vi = 0m/s
t =?

Since you can't solve for t, i solved for Vf.

Would the Vf be a negative number?

Then, i pluged the vi and vf into the 2nd equatioon above, but the answer i got was different from the answer given.
 
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Why can't you solve for t?
 
o wait, maybe i made a mistake.
 
I forgot that my Vi = 0 m/s therefore does not equate to a quadratic.
 
Well, use the first equation you wrote to solve for t. You know all the information you need to solve for part a.

The D = -500 m a = -g t=? and vi= 0 The reason why I say there is no initial velocity is because once the car goes over the edge and starts to fall, it's in free fall. The car may have HORIZONTAL velocity, but there is no initial vertical velocity.

Use that information and you can solve for v final.
 
Even if it's quadratic, why wouldn't you be able to solve for t?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

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