A ball is dropped from a tower of height 80 ~m . The time it takes to cover the last 50 % of its fall is (acceleration due to gravity =10 ~ms ⁻² )

Step-by-step solution

The given situation is shown below For Ist part of journey, aligned & u=0, a=-g=-10 ~ms ⁻² & s=-40 ~m aligned Now using, v^2-u^2=2 a s we get, v^2-0=2 (-10)(-40) v^2=800 ...(i) And using v=u+a t , we get; v=0+a t₁ , or v^2=a^2 t₁^2 t₁^2= v^2 a^2 = 800 100 =8 t₁=2 2 Now, for second part of journey, u=0, a=-10 ~m / s ^2, s=-80 ~m v^2-u^2=2 a s v^2=2 (-10)(-80)=1600 And v=u+a t₂ gives, v=a t₂ v^2=a^2 t₂^2 Hence, 1600=100 t₂^2 or t₂=4 ~s Time to complete last 50 % part of the journey =t₂-t₁=4-2 2 1.17 ~s