Problem 1 The vertical distance travelled by an object under gravity and with ai

Question

Problem 1 The vertical distance travelled by an object under gravity and with air resistance is given by 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,0.15 kg/sec, and g-9.81 m/s, 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 per- centage error for each iteration. Carry out the problem with five iterations or four significant digits, whichever comes first : :39t 80(1-e-o,0125t

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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