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15 votes
15 votes
A body dropped from rest falls half its total distance in the last second before it strikes the ground.From what height was it released

User Anujprashar
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2.8k points

2 Answers

25 votes
25 votes

Answer:

it is dropped from a height of 10m

Step-by-step explanation:

User Binta
by
2.7k points
12 votes
12 votes

Answer:

57.1 meters

Step-by-step explanation:

the distance the body has fallen is given by

d = 1/2 a t^2 with a = 9.8m/s^2 this becomes

d = 4.9 t^2

now we want the distance in one more second to equal the distance fallen before that one second

or 4.9 t^2 = 4.9 (t+1 )^2 - 4.9 t^2 re-arrange

2 * 4.9 t^2 = 4.9 (t+1)^2 simplify to

0 = - 4.9 t^2 + 9.8 t + 4.9 use Quadratic Formula to find

t = 2.414

then t +1 = 3.414

this corresponds to distance d = 1/2 at^2 = 1/2 ( 9.8) (3.414)^2 = 57.1 meters

check: in 2.414 seconds distance = 1/2 a t^2 = 28.6

in one second more distance = 1/2 (9.8)(3.414)^2 = 57.1

so in the last one second it has fallen approx another 28.6 m

Check !

User Shimon S
by
2.8k points