A body dropped from a height h falls half of its distance in the last second
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Question A body is dropped from height h. If the body covers the distance of h/3 in the last second, what was the total height?
Answer
d=1/2 at²
Let (t-1) be the distance covered in the first 2/3rds of motion. Then:
2d/3=1/2 a(t-1)²
d=1/2 at²
2d/3 /d=1/2a(t-1)² / 1/2 a t²
2/3=(t-1)² / t²
3(t-1)²=2t²
3t²-6t+3=2t²
t²-6t+3=0
t=5.4495 secs
h(5.4495)=145.515m
Answer
d=1/2 at²
Let (t-1) be the distance covered in the first 2/3rds of motion. Then:
2d/3=1/2 a(t-1)²
d=1/2 at²
2d/3 /d=1/2a(t-1)² / 1/2 a t²
2/3=(t-1)² / t²
3(t-1)²=2t²
3t²-6t+3=2t²
t²-6t+3=0
t=5.4495 secs
h(5.4495)=145.515m
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