Question

A rock is thrown at an angle of 30 degrees above the horizontal with initial velocity 15m/s what is the displacement when the rock returns to the ground

Answer 1

The displacement of the rock will be the same as the total horizontal distance traveled. Here the rock's horizontal position is given by

$x=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\cos30^\circ\,t$

so to find the horizontal distance it traversed, we need to know the time it took for the rock to return to the ground. We use the rock's vertical position over time to figure that out:

$y=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\sin30^\circ\,t-\dfrac g2t^2=0$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the acceleration due to gravity. Then we find that $t\approx1.5\,\mathrm s$, at which point we find $x\approx20\,\mathrm m$.