Question

A ball is projected with velocity 10√2 at an angle of 45 degree with the horizontal from the top of a cliff 20 m high with what velocity does the ball hit the ground​

Answer 1

Answer:10$\sqrt{6}$

Explanation:    

we know, u=10$\sqrt{2}$

       Ux=10$\sqrt{2}$cosθ=10√2cos45=10m/s

       Uy=10$\sqrt{2}$sinθ=10√2sin45=10m/s

          Vx=Ux=10m/s  (at all points)

          Sy= Uyt-$\frac{1}{2}$gt²

         -45=10t-$\frac{1}{2}$10t²

           t=1+√5 sec

           Vy=Uy-gt

               =10-10(1+√5)

              Vy =-10√5 m/s

           V²=Vx²+Vy²

           V=$\sqrt{Vx^{2} +Vy^{2} }$=$\sqrt{10^{2}+10\sqrt{5} ^{2} }$  =10$\sqrt{6}$