6i think the ans is ga2 / b + √a2 + b2 indeed
What is the minimum speed to hit the co-ordinate (a,b) if fired from the origin ? (gravity = g , along negative Y axis)
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11 Answers
1is this the answer???
v_{min} = \frac{ga^{2}}{b + \sqrt{a^{2}+ b^{2}}}
49are u sure that the terms will be irrespective of the u term?
i think that cannot be the case!
i get this as answer:
v_{min}=\sqrt{\left(u-1 \right)^2+\left( \frac{ag}{u}\right)^2}
u=initial vel of projection
11st with the help of equation of trajectory we can write...
b= a tan\phi - ga22u2cos2 \phi → (1)
in this equation there are 2 unknown namely , \phi and u.
in the question its given that it shud reach (a,b) with min. speed.... it means that at b final speed should be zero.....so applying that condition...
0= u2sin2\phi - 2gb →(2)
so there are 2 unknown and 2 equation....solving these we should get the ans......
1subhomoy 'umin' hi toh nikalna hai
i too made a mistake in posting the above answer
v_{min} = \sqrt{\frac{ga^{2}}{b + \sqrt{a^{2}+b^{2}}}}
i solved it like this:
from -equation-of-trajectory
v^{2} = \frac{1}{2}\frac{ga^{2}}{(atan\Theta - b)cos^{2}\Theta}
so for vmin we have to maximise - (atanθ - b)cos2θ
using application of derivative ,for maximum value of above term
tan\Theta = \frac{b + \sqrt{a^{2} + b^{2}}}{a}
and
2cos^{2}\Theta = \frac{b +\sqrt{a^{2} + b^{2}}}{\sqrt{a^{2} + b^{2}}}
putting these values in first equation
v_{min} = \sqrt{\frac{ga^{2}}{b + \sqrt{a^{2}+b^{2}}}}
49i used,
a=ucosθt
and found value of t in terms of a
vx=ucosθ
vy=usinθ-gt=usinθ-agucosθ
then used v=√vx2+vy2
and applied maxima and minima for fixed u and variable θ!!
1from the equation of trajectory
b = a tanθ - ga2/2v2cos2θ
b = atanθ - ga2(1+tan2θ)/2v2
(ga2/2v2)tan2θ - atanθ + (ga2/2v2+b) = 0
tanθ is real ,
a 2 >= 4.(ga2/2v2)(ga2/2v2+b)
(v4 - 2bgv2 ) ≥ g2a2
(v2 - bg)2 ≥g2(a2+b2)
Vmin = √bg+g√(a2+b2)