A particle is projected from ground ,find the maximum angale of projection so that particle always seems to be moving away from the point of projection?
(ONLY FOR IIT JEE PREPARATION STUDENTS/IITAINS)
plz help me
IG-karsiddharth7
Answers
Answer:I am guessing what your question really meant is “What is the maximum angle of launch so that the total DISTANCE of the projectile from the launch point is ALWAYS increasing.” Obviously at a very steep angle, the projectile can travel some distance, but fall back very close to the source. The horizontal distance x is given by
x=cosθ⋅v⋅t
and the vertical distance by
y=sinθ⋅v⋅t−(g/2)⋅t2
where v is velocity, g the gravitational constant and θ
is the launch angle from the horizontal.
The total distance from the origin is
D=√(x2+y2)=√(v2⋅t2+(g/2)2⋅t4−g⋅t3⋅v⋅sinθ)
For this distance D to be constantly increasing, simply make sure that the derivative dD/dt>0
Working through the algebra this results in:
g2⋅t2−3g⋅t⋅v⋅sinθ+2⋅v2>=0
This is a quadratic equation in t . It is easy to see that for the solution be real
(9sin2θ−8)>0
Hence
θ = 70.529 deg.
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