Question

A projectile shot at an angle of 60 degrees above the horizontal ground strikes a vertical wall 30m away at a point 15m above the ground. Find the speed with which the projectile was launched and the speed with which it strikes the wall. sorry experts the previous one is the wrong one.

Answer 1

The equation of trajectory of projectile is give by:

$$\begin{gathered} y=x\times \tan \theta -\frac{x^2\times g}{2u^2{\cos }^2\theta } \\ substitutingthevaluesfromthequestion \\ x=30m \\ y=15m \\ \theta ={60}^{\circ } \\ andg=10m/s^2 \\ 15=30\tan {60}^{\circ }-\frac{{30}^2\times 10}{2u^2\times 0.5^2} \\ u=22.07m/s \\ {u}_x=u\cos 60=11.035m/s \\ {u}_y=u\sin 60=19.11m/s \\ Maximumheight \\ H=\left(\frac{u^2{\sin }^2\theta }{2g}\right)=18.26m \\ dis\tan cefellbytheprojectile=h=H-15=3.26m \\ Timeofflight=t=\frac{2u\sin \theta }{g}=3.8226s \\ timewhenprojectileisathighestpoint=t/2=1.9113swhichmeansprojectilehascrossedthehalfwaypoint. \\ nowverticalvelocityatthispoint{v_y}^{\text{'}}=\sqrt{2gh}=8.074m/s \\ horizontalvelocityatthispoint{v_x}^{\text{'}}=v_x(\sin cejorizontalvelocitydoesnotchangeforaprojectile) \\ thereforevelocityatthispoint,i.e.,whenithitsthewall=v^{\text{'}}=\sqrt{{\left({v_y}^{\text{'}}\right)}^2+{\left({v_x}^{\text{'}}\right)}^2}=\sqrt{8.{074}^2+11.{035}^2}=13.673m/s \end{gathered}$$