A cat leaps horizontally from a flat roof 7.5 m above the ground and lands on another flat roof 5 m above the ground and 3.5 m away horizontally.
how long in seconds does the leap take?
2007-06-14 01:45:38 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The first two answers are correct but there was no mention of neglecting air resistance so the answer would be somwat longer.
2007-06-14 02:27:10 · answer #1 · answered by confused 3 · 0⤊ 0⤋
the time taken to in seconds is as follows
As the cat leaps horizontally its vertical velocity initially is zero hence time taken is given by the formula
S = u * t + .5 * a * t * t
S= distance travelled = 7.5-5 = 2.5m
u = initia velocity = 0
a= g = 9.81 m/s*s
hence t = (2.5 * 2 / 9.81)^.5
= .7139 sec
2007-06-14 08:53:01 · answer #2 · answered by shrikant s 2 · 1⤊ 0⤋
The cat has fallen 2.5 m and
s = at²/2 where s = distance and a = 9.8 m/s². So
t = â(2s/a) = â(2*2.5/9.8) = .714 s.
Doug
2007-06-14 08:58:34 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋