how fast would an object be travling after falling for 35 ft?
2006-08-28 06:18:42 · 8 answers · asked by mr._693 3 in Science & Mathematics ➔ Mathematics
do the math. find a thirty five foot drop and time from the point of release to the point of impact. use the acceleration formula- 9.8 meters per second squared.
2006-08-28 06:22:20 · answer #1 · answered by SST 6 · 0⤊ 1⤋
I guess you are talking about an object falling on the surface of the Earth, not one in Space.
On Earth the strength of gravity is an equation about acceleration. It means that an object will accelerate 32 feet per second, per second.
After falling 35 feet, the object would have been falling for just over 1 second. It would have accelerated 3/32 of 32 feet per second faster than the 32 feet per second that it achieved in the first 1 second.
Its speed is now 35 feet per second. In 2 seconds after its release, it will be traveling 64 feet per second.
;-D Sorry for not doing it in algebra. I hate math!
2006-08-28 13:30:34 · answer #2 · answered by China Jon 6 · 0⤊ 0⤋
It would be 9.81 m per second (square) or you could approximate it to 10 m. (normal resistance)
If the object is offering increased or decreased resistance while travelling down the the speed can be increased or decreased by exposing the body surface that offers the resistance just like sky-divers do.
http://www.antonine-education.co.uk/Physics_AS/Module_2/Topic_1/free_fall_and_terminal_speed.htm
2006-08-28 13:26:02 · answer #3 · answered by Bramhastra 3 · 1⤊ 0⤋
Free fall means no air resistance. So we have:
h = g*t^2/2 => t = sqrt(2h/g)
v = g*t => v = g*sqrt(2h/g) => v = sqrt(2*h*g)
where h = 35 ft and g = 32.2 ft/sec^2
so
v = sqrt(2*35*32.2) = 47.5 ft/sec
2006-08-28 13:37:45 · answer #4 · answered by Dimos F 4 · 0⤊ 0⤋
Is it in a vacuum? If not it depends on the terminal velocity of the object.
2006-08-28 13:25:05 · answer #5 · answered by N3WJL 5 · 0⤊ 0⤋
I am not so familiar with english units so I will try to solve your problem in SI..
35 ft x 12in/1ft x 2.54cm/in x 1m/100 cm= 10.668m
s=0.5 (g)t^2
t= sqrt (10.668/(0.5*9.81)
t= 1.47 sec
a= (vf-vi)/t
vi=0m/sec
9.81=vf/1.47
vf=14.42 m/sec or 47.31 ft/sec
2006-08-28 13:34:41 · answer #6 · answered by cooler 2 · 0⤊ 0⤋
v^2=2*32*35
v=approximately 48'/sec
2006-08-28 13:33:18 · answer #7 · answered by raj 7 · 0⤊ 0⤋
final velocity = sqrt(2 * g * d)
=sqrt( [2] * [32](ft/sec^2) * [35](ft) )
=sqrt( [2240] (ft/sec)^2 )
=47 ft/sec
2006-08-28 20:51:56 · answer #8 · answered by Anonymous · 0⤊ 0⤋