Physic Labs

Problem 1

A ball is launched horizontally from a cliff of height h=45 mh=45\,\mathrm m with initial speed u=10 m s−1u=10\,\mathrm{m\,s^{-1}}. Neglect air resistance and take g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Choose the launch point as the origin, with the horizontal axis forward and the vertical axis upward. The ground is a vertical distance hh below the launch point; treat the ball as a point particle that hits no other object before landing. Measure time from the instant it leaves the cliff. (a) Find the time to reach the ground and the horizontal range. (b) Determine the velocity components just before impact. (c) Find the impact speed and its angle below the horizontal.
t=2hg=3.0 st=\sqrt{\frac{2h}{g}}=3.0\,\mathrm s
problems.proof.analysis

The vertical displacement to the ground is hh, so the free-fall equation determines the flight time. The positive root is chosen because time after launch is positive.

problems.proof.pitfall. Do not assign the same vertical and horizontal velocity: horizontal motion is uniform, while the vertical component is accelerated by gravity.