The vertical distance travelled by an object under gravity and with air resistan
ID: 1842514 • Letter: T
Question
The vertical distance travelled by an object under gravity and with air resistance is given by s = mg/k t - m^2/k^2 g (1 - e^k/m t)_t where m is the mass of the object, g is the acceleration due to gravity, k is the damping constant, s is the distance, and t is the time. Given that m = 12 kg, k = 0.15 kg/sec, and g = 9.81 m/s^2, find out the time required for the object to travel 300 m. Use the fixed-point iteration method with an initial guess equal to the time required by the object to travel 300 m. Neglect air resistance Determine the relative percentage error for each iteration. Carry out the problem with five iterations or four significant digits, whichever comes first.Explanation / Answer
solution:
1)here we have to solve above problem by fixed point iteration or successive approximation method as follows
2)where time required for object to travel 300 m is given by law of motion as
s=ut+.5*gt^2
u=0
s=.5gt^2
300=.5*9.81*t^2
t=7.8206 sec
3)here equation is given by
y=f(t)=0
s=(mg/k)t-(m/k)^2*g*[1-e^(-kt/m)]
on putting value we get equation as
784.8t-62784e^(-.0125t)-63084=0
for this method we put it as
t=g(t)
t=80e^(-.0125t)+80.382263
4)so here equation becomes for
t=80e^(-.0125t)+80.382263
initial guess,t=7.8206 sec
gives
t1=152.93177 sec
percentage error=[(t1-t)/t]*100
P.E.=1855.49%
5)for second iteration
so here equation becomes for
t=80e^(-.0125t)+80.382263
initial guess,t=152.93177 sec
gives
t1=92.2092 sec
percentage error=[(t1-t)/t]*100
P.E.=39.70566%
6)for third iteration we get that
here equation becomes for
t=80e^(-.0125t)+80.382263
initial guess,t=92.2092 sec
gives
t1=105.64705 sec
percentage error=[(t1-t)/t]*100
P.E.=14.57322%
7)here for fourth iteration we get that
so here equation becomes for
t=80e^(-.0125t)+80.382263
initial guess,t=105.64705 sec
gives
t1=101.74052 sec
percentage error=[(t1-t)/t]*100
P.E.=3.69772%
8)for fifth iteration we get that
so here equation becomes for
t=80e^(-.0125t)+80.382263
initial guess,t=101.74052 sec
gives
t1=102.8093 sec
percentage error=[(t1-t)/t]*100
P.E.=1.0505%
9)for sixth iteration we get that
so here equation becomes for
t=80e^(-.0125t)+80.382263
initial guess,t=102.8093 sec
gives
t1=102.51172 sec
percentage error=[(t1-t)/t]*100
P.E.=.28951%
10)on proceeding in same way we get accurate answer for four significat figure at 13 th iteration as
so here equation becomes for
t=80e^(-.0125t)+80.382263
initial guess of previous 12 th iteration,t=102.57626 sec
gives
t1=102.57629 sec
percentage error=[(t1-t)/t]*100
P.E.=.00003%
11)hence our answer is t=102.57629 sec
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