A 6m ladder leans against a wall. The bottom of the ladder is 1.3m from the wall at time =0sec and slides away from the wall at a rate of 0.3m/s. Find the velocity of the top of the ladder at time =2 (take the direction upwards as positive).(Use decimal notation. Give your answer to three decimal places.)

Answer:- 0.100 Step-by-step explanation:Length of the ladder,  H = 6 mDistance at the bottom from the wall, B = 1.3 mLet the distance of top of the ladder from the bottom at the wall is PThus,from  Pythagoras theorem,B² + P² = H²    .orB² + P² = 6²          ..............(1)      [Since length of the ladder remains constant]at B = 1.3 m1.3² + P² = 6²orP² = 36 - 1.69orP² = 34.31orP = 5.857 Now,differentiating (1)[tex]2B(\frac{dB}{dt})+2P(\frac{dP}{dt})=0[/tex]at t = 2 secondschange in B = 0.3 × 2= 0.6 ftThus,at 2 secondsB = 1.3 + 0.6 = 1.9 mtherefore,1.9² + P² = 6²orP = 5.69 mon substituting the given values,2(1.9)(0.3) + 2(5.69) × [tex](\frac{dP}{dt})[/tex] = 0or1.14 + 11.38 × [tex](\frac{dP}{dt})[/tex] = 0or11.38 × [tex](\frac{dP}{dt})[/tex] = - 1.14or [tex](\frac{dP}{dt})[/tex] = - 0.100 here, negative sign means that the velocity is in downward direction as upward is positive