Asked by Ryzza
If a compact body falls vertically 16 ft. during the 1st second, 48 ft. during the 2nd second, and 80 ft. during the third second, how far will it fall during the 7th second? During the first 7 seconds?
Answers
Answered by
Reiny
total distance fallen = 16t^2 , where t is in seconds
after 1 second --- 16(1)^2 = 16
after 2 seconds --- 16(2^2) = 48
after 3 seconds --- 16(3^2) = 144
Where did you get your data from, it certainly is not from this earth.
But .... ignoring the silliness of your data,
you have an arithmetic series
a = 16 , d = 32
term(7) = a + 6d = 16+6(32) = 208
sum(7) = (7/2)(16 + 208) = 784
after 1 second --- 16(1)^2 = 16
after 2 seconds --- 16(2^2) = 48
after 3 seconds --- 16(3^2) = 144
Where did you get your data from, it certainly is not from this earth.
But .... ignoring the silliness of your data,
you have an arithmetic series
a = 16 , d = 32
term(7) = a + 6d = 16+6(32) = 208
sum(7) = (7/2)(16 + 208) = 784
Answered by
Anonymous
208
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