A car smashes through a safety rail, landing 50m from a 100m high vertical cliff...

topsquark

Forum Staff
Apr 2008
3,115
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On the dance floor, baby!
I see you found the site okay. :)

I would first sketch the problem, then set up a coordinate system.

I will use the usual coordinate directions: +x to the right and +y upward. I am going to set an origin at the base of the cliff: 100 m directly below where the car leaves the ground. I am taking the initial velocity of the car to be in the +x direction.

a) We know that the horizontal and vertical parts of the motion are independent of one another, and that the car has an initial velocity component in the y direction of 0 m/s. So
\(\displaystyle y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2\)

\(\displaystyle 0 = 100 + \frac{1}{2}(-9.8)t^2\)
Solve for t.

b) We know that the car's initial velocity was directed in the x direction. We know how long it takes for the car to hit the ground from part a). And we know that there is no acceleration in the x direction. Thus
\(\displaystyle x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2\)

\(\displaystyle 50 = v_0t\)
Solve for \(\displaystyle v_0\).

c) We are asked for the velocity so we need to find the vector sum of the car's final velocity components in the x and y directions. We already know the x component from b), since the x component of the velocity never changes. We know how long it took the car to fall from a), so
\(\displaystyle v_y = v_{0y} + a_yt\)

\(\displaystyle v_y = -9.8t\)

Now add \(\displaystyle v_x\) and \(\displaystyle v_y\) as vectors.

-Dan