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

#### topsquark

Forum Staff
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