Projectile Motion Q (1 Viewer)

[shaon0]

Two stones are thrown simultaneously from the same point at ground level in the same direction and with the same non-zero angle of projection @, but with different velocities, U, V metres per second, where U < V. The slower stone hits the ground at point P. At that instant the faster stone just clears a wall of height h metres above ground llevel with an angle of B to the horizontal.

i) Show that, while both stones are in flight, the line joining them has an inclination to the horizontal which is indepedent of time.

ii) Show that V(tan@+tanB)=2Utan@

iii) If B=@/2, deduce that U < (3V/4)

Thanks for the help.

 

[azureus88]

[maths]x_1=Ut\cos\alpha \\y_1=-\frac{g}{2}t^2+Ut\sin\alpha \\x_2=Vt\cos\alpha \\y_2=-\frac{g}{2}t^2+Vt\sin\alpha \\\\m=\frac{y_2-y_1}{x_2-x_1}=\frac{t\sin\alpha (V-U)}{t\cos\alpha (V-U)}=\tan\alpha \\\therefore $Angle of inclination$\,=\alpha \\\\$Let T be time of flight for particle 1$\\T=\frac{2U\sin\alpha }{g}\\$at t=T,\\\tan\beta =-(\frac{-gT+V\sin\alpha }{V\cos\alpha })\\V\cos\alpha \tan\beta =2U\sin\alpha -V\sin\alpha \\V\tan\beta =2U\tan\alpha -V\tan\alpha\\\therefore V(\tan\alpha +\tan\beta )=2U\tan\alpha \\\\$Let$\,\, t=\tan\frac{\alpha }{2}\\V(\frac{2t}{1-t^2}+t)=2U(\frac{2t}{1-t^2})\\V(3-t^2)=4U\\U=\frac{V}{4}(3-t^2)\\$since$\,\,0< \alpha < \frac{\pi}{2},\,\,0< t^2< 1\\\therefore \frac{V}{2}< U< \frac{3V}{4}[/maths]

 

[shaon0]

Thanks