how long to the nearest tenth will the ball fall 225ft?
s = 16t^(2) + 32t, s is in feet.
2006-06-05 13:55:17 · 5 answers · asked by ikaryuu 1 in Science & Mathematics ➔ Mathematics
Set 16t^2 +32t -225=0
Use quadratic formula gives t = 2.9sec
2006-06-05 14:43:26 · answer #1 · answered by Jimbo 5 · 0⤊ 0⤋
if the ball is at rest, distance s = 16t^(2)
if it is already falling, then
s = 16t^(2) + (initial velocity)*t
if you say,
s = 16t^(2) + 32t
this means that the initial velocity is 32 ft/sec
in which case
225 = 16t^(2) + 32t
subtract 225 from both sides
0 = 16t^(2) + 32t - 225
need to use quadratic equation (http://www.1728.com/quadratc.htm) where
a = 16, b = 32, and c = -225
and i get a time of 2.88 seconds
2006-06-05 21:43:33 · answer #2 · answered by ledezmajr 3 · 0⤊ 0⤋
Yeah, 2.9 sec, but realize that your equation is NOT for a ball simply dropped, but rather one that is given an initial downward velocity of 32 ft/sec.
If you meant to release it with no velocity (Vo=0), the eq of motion is just s=16t^2, so t would be 3.75 sec.
2006-06-05 21:48:40 · answer #3 · answered by Steve 7 · 0⤊ 0⤋
s = 16t^(2) + 32t
225 = 16t^(2) + 32t
16t^(2) + 32t - 225 = 0
t = [-32 +- SQRT(15424)]/(2)(16)
t = 2.88 or -4.88 (NA)
t = 2.9 s (to the nearest tenth)
2006-06-05 23:45:26 · answer #4 · answered by Kemmy 6 · 0⤊ 0⤋
about 2.9 seconds
2006-06-05 21:43:27 · answer #5 · answered by davidosterberg1 6 · 0⤊ 0⤋