a ball is dropped from the height. if it takes 0.2 sec. to cross the last 6 m before hitting the ground find the ht. from which it was dropped

Answer 1

(Time to reach ground) - (Time to reach 6 m above ground) = 0.2 s

sqrt(2H/g) - sqrt[2(H-6)/g] = 0.2

Let g = 9.8 m/s^2 and solve for H

0.4517[sqrtH - sqrt(H-6)] = 0.2
sqrtH - sqrt(H-6) = 0.4427

H = 49 m