Asked by Anonymous
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
Answers
Answered by
drwls
(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
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
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