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3-2.Motion in Plane
medium
A cannon on a level plane is aimed at an angle $\theta $ above the horizontal and a shell is fired with a muzzle velocity ${v_0}$ towards a vertical cliff a distance $D$ away. Then the height from the bottom at which the shell strikes the side walls of the cliff is
A
$D\sin \theta - \frac{{g{D^2}}}{{2v_0^2{{\sin }^2}\theta }}$
B
$D\cos \theta - \frac{{g{D^2}}}{{2v_0^2{{\cos }^2}\theta }}$
C
$D\tan \theta - \frac{{g{D^2}}}{{2v_0^2{{\cos }^2}\theta }}$
D
$D\tan \theta - \frac{{g{D^2}}}{{2v_0^2{{\sin }^2}\theta }}$
Solution
(c) Equation of trajectory for oblique projectile motion
$y = x\tan \theta – \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}$
Substituting $x = D$ and $u = {v_0}$
$h = D\tan \theta – \frac{{g{D^2}}}{{2u_0^2{{\cos }^2}\theta }}.$
Standard 11
Physics
