Problem 1

A small object is launched horizontally from the edge of a cliff of height

H

above level ground with initial speed

v_0

. Set the origin at launch,

x

horizontal along the launch direction and

y

upward; neglect air resistance and use

g=9.8\,\mathrm{m/s^2}

. The ground is level and wide enough to catch it. (a) Find flight time and range

R

. (b) Find the velocity components and speed just before impact. (c) Determine the trajectory angle below the horizontal at impact.

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

\tan\phi=\frac{|v_y|}{v_x}=\frac{\sqrt{2gH}}{v_0},\quad \boxed{\phi=\arctan\frac{\sqrt{2gH}}{v_0}}

problems.proof.analysis

The angle below horizontal follows from the ratio of downward to horizontal velocity. Together, $t_f=\sqrt{2H/g}$ , $R=v_0t_f$ , with the impact components and speed above.

problems.proof.pitfall. Do not set

v_y=0

at impact: only the initial vertical velocity is zero. The final vertical component is negative with upward-positive coordinates.