An object is droppedfrom a height h from the ground. Every time it it hitsgroundit loses50 %of its kineticenergy . The total distancecoveredast to oo is :

An object is droppedfrom a height h from the ground. Every time it it

1.	An object is droppedfrom a height h from the ground. Every time it it hitsgroundit loses50 %of its kineticenergy . The total distancecoveredast to oo is :
Answer» 3H `oo` `5/3` h `8/3` h Solution :LET mass of object is m , and dropedfrom heighth on the GROUND, energyis convertedfrom P.E toK.E Kinetic energy of object = potentialenergy of object . K.E = mg(h) Thisobjecthitethe groundbut loose50 % ofinitial K.E K.E =mg(h/2) Every time object halfof the heightfrom which it was DROPPED. Total distancecovered by object hiting the ground toll `t to oo` `H = h+2 (h/2 +h/(2^(2)) +....+h/(2^(n))) = h+h (1+1/2 +1/(2^(2))+....oo)` `= h + h (1/(1-1/2))` H = 3 h