Please show equations to work this out! Cheers!
2006-09-28 01:08:38 · 13 answers · asked by W J 1 in Science & Mathematics ➔ Physics
hm... the weight of the stone won't even matter, the size will...
2006-09-28 01:09:37 · answer #1 · answered by Walter W. Krijthe 4 · 0⤊ 2⤋
If it is thrown vertically use: v^2 =u^2 + 2as:
take upwards as the positive direction.
u = initial velocity = 8 ms^-1
v = final velocity = 0 ms^-1 at max height
a = acceleration = -g (acceleration due to gravity which acts in tne -'ve direction) = -9.8 ms^-2
Hence: 0 = 8^2 + (2x -9.8 x s)
0 = 64 -19.6s
19.6s = 64
s = 64/19.6
s = 3.2653
s = 3 .27 m (3sf)
2006-09-28 08:23:21 · answer #2 · answered by RATTY 7 · 0⤊ 0⤋
the maximum height gained is 3.2 meters
initial velocity=8 m/s
final velocity(v)=0m/s(because the stone is thrown vertically up)
-(u)2=-2gh
h=(u square)/2g=8*8/2*10=64/20=3.2meters
2006-09-28 08:31:52 · answer #3 · answered by faraday 1 · 0⤊ 0⤋
yeah do your own homework. Anyway, we guess you threw it straight up. I can't recall the formula: is it
v^2 = u^2 + 2 * a * s
where
u is the initial velocity
v is the final velocity (zero, cos that's the point when it's highest)
a is acceleration (gravity will do)
and s is the distance it has travelled
Solve for s and you've got it. IF i remembered it right...
2006-09-28 08:15:39 · answer #4 · answered by wild_eep 6 · 0⤊ 0⤋
Max height = straight up
Force to beat = gravity = mg
m = mass
g = gravity
Force applied = initial speed = ma = 0.5 m v^2
Notice how I have not given you h = height
work that into hte equation, plug in numbers and you will have learned something.
2006-09-28 08:18:25 · answer #5 · answered by Anonymous · 0⤊ 0⤋
aint got a 4 by 2
2006-09-28 08:15:59 · answer #6 · answered by FLOYD 6 · 0⤊ 0⤋
v=8met/sec, u=0 Assume g=9.81met/sec^2
Assume also that stone is projected vrtically upwards
v^2-u^2=2gs
64=2*9.81*s
s=64/2*9.81=3.26met
2006-09-29 13:35:03 · answer #7 · answered by helenagilchrist@btinternet.com 1 · 0⤊ 0⤋
v=8met/sec, u=0 Assume g=9.81met/sec^2
Assume also that stone is projected vrtically upwards
v^2-u^2=2gs
64=2*9.81*s
s=64/2*9.81=3.26met
2006-09-28 08:22:36 · answer #8 · answered by openpsychy 6 · 0⤊ 0⤋
Wouldn't it depend on the angle you threw the stone?
I think you should do your own homework realy!
2006-09-28 08:10:54 · answer #9 · answered by Alonsofan 3 · 0⤊ 0⤋
the answer is 1.9meters above ground level but as soon as it leaves whatever its being thrown by the force of gravity takes it down i think its 9.8neutons
2006-09-28 08:31:03 · answer #10 · answered by GHIVE 2 · 0⤊ 0⤋
the max height is below the earths atmosphere
2006-09-28 08:16:12 · answer #11 · answered by Anonymous · 0⤊ 0⤋