CBSE Class 11-science Answered
it is a projectile motion based , i think a tough question , not getting answer even by several attempt
Asked by | 24 Apr, 2012, 04:38: PM
Expert Answer
The range of the projectile motion is given by
R=u2sin2?/g
Time of flight is given by
T=2usin?/g
Let d? be the fractional change in the value of the angle.
Differentiating R w.r.t ? we get
dR/d? = 2u2cos2?/g
Multiply by d? on both sides, we get
dR= 2u2cos2?d?/g
This represents the change in range, so fractional change in range would be to divide this change with the original value of range.
Fractional change in Range = dR/R = 2cos2?d?/sin2? =x
Now,
Differentiating T w.r.t ? we get
dT/d? = 2ucos?/g
Multiply by d? on both sides, we get
dR= 2ucos?d?/g
This represents the change in time of flight, so fractional change in time of flight would be to divide this change with the original value of time of flight.
Fractional change in Time of flight = dT/T = cos?d?/sin? =y
So, x= 2cos2?d?/sin2? and y= cos?d?/sin?
And ?=60, putting the value of ? we get
x=2(cos120)d?/sin120 = -2d?/?3
y=cos60d?/sin60 = d?/?3 = -x/2 is the answer
Answered by | 30 Apr, 2012, 03:06: PM
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