A cannon ball is fired from the top of a 55 ~m high cliff with an initial speed of 50 ~m ~s ⁻¹ . The speed of the cannon ball while hitting the ground in m s ⁻¹ (acceleration due to. gravity =10 ~ms ⁻² )

Step-by-step solution

The given situation is shown below According to work-energy theorern, Initial K. E of cannon ball + Gravitational P.E = Total K. E at ground 1 2 m v₁^2+m g h= 1 2 m v₂^2 v₁^2+2 g h=v₂^2 v₂^2=v₁^2+2 g h Here, v₁= initial speed =50 ~m / s h= height of cliff =55 ~m So, v₂^2=(50)^2+(2 10 55) aligned & v₂^2=2500+1100=3600 & or v₂= 3600 =60 ~m / s aligned