a)
Definitely there is a
pattern in the graph. A stone projected upwards , reaches greatest height 125 (appr)
at around 5 seconds and then begins to fall. If we draw a vertical from the vertex
(greatest height of 125 around) , then the curve is symmetrical about this vertical
line. The the height travelled in every succesive equal interval is diminishing till the
greatest height is reached and then it increases while
falling.
The values taken below is from the graph by the
eye judgement and so is approximate
only.
b)
At t=2 and t = 8, the
graph is very clearly near round figure.
For us it looks
that at t = 2, H(2) = 79 and at t = 8 , H(8) = 81.
Also we
know that H(t) = ut-(1/2)gt^2 , where u is the initial velocity , g is the acceleration
due to gravitation.
We estimate the u and g from the
graphical positions of (t, H(t)) at t = 2 and t =8 which looks better accurate points
from visual accuracy.
H(t) =
ut-(1/2)t^2.
H(2) = 2u -2g =
79.......(1).
H(8) = 8u -32g =
81......(2).
16eq(1) - eq(2) gives 16(2u-2g)-(8u-32g) =
79*16-81 = 1183
24u =
1055
u = 105/24 = 49.2917
m/s
4(1) - (2) gives 4(2u-2g)-(8u-32g) =
79*4-81= 235
24g =
235
g = 235/24 = 9.7917
m/s^2.
Therefore , H(t) = 49.2917t -
(1/2)9.7917t^2 is the equation.
From this , when H(t) = 0.
Or 49.2917t - (1/2)9.7917 t^2 = 0
t(49.2917 -
(1/2)(9.7917)t) = 0
So t = 0,
or
49.2917 - (1/2)9.7917t = 0. or t = 2*49.2917/9.7917 =
10.068 secs approximately.
Hope this
helps.
No comments:
Post a Comment