Originally Posted by

**Chris L T521** At $\displaystyle t=0$, the stone is thrown horizontally off the cliff with the following velocities: $\displaystyle v_{0x}=15\,m/s,\,v_{0y}=0\,m/s$. Let $\displaystyle g=9.8\,m/s^2$, $\displaystyle y_0=30\,m$, $\displaystyle y_f=0\,m$, $\displaystyle x_0=0\,m$ and $\displaystyle x_f=?$

From Kinematics, we know that $\displaystyle \Delta y=v_{0y}t-\tfrac{1}{2}gt^2\implies -30m=\left(-4.9\,m/s^2\right)t^2\implies t^2=6.12\,s^2\implies t\approx 2.47\,s$

We also know that $\displaystyle \Delta x= v_{0x}t=15\, m/s\cdot 2.47\,s=37.11\,m$

Thus, the stone travels $\displaystyle 37.11\,m$ in the horizontal direction, doing so in $\displaystyle 2.47\,s$

Does this make sense?