Hard
Question
A ball is thrown eastward across level ground. A wind blows horizontally to the east, and assume that the effect of wind is to provide a constant force towards the east, equal in magnitude to the weight of the ball. The angle θ (with horizontal) at which the ball should be projected so that it travels maximum horizontal distance is
Options
A.45o
B.37o
C.53o
D.67.5o
Solution
Since time of flight depends only on vertical component of velocity and acceleration . Hence time of flight is
T =
where ux = cosθ and uy = u sinθ
∴ In horizontal (x) direction
d = uxt + 1/2 gt2 = u cosθ
=
(sinθ cos θ + sin2θ)
We want to maximize f(θ) = cosθ sinθ + sin2θ
⇒ f ′θ = − sin2θ + cos2θ + 2 sinθ cosθ = 0
⇒ cos2θ + sin2θ = 0
⇒ tan2θ = −1
or 2θ =
or θ = 
= 67.5o
Alternate :
As shown in figure, the net acceleration of projectile makes on angle 45o with horizontal. For maximum range on horizontal plane, the angle of projection should be along angle bisector of horizontal and opposite direction of net acceleration of projectile.
∴ θ =
= 67.5o
T =
where ux = cosθ and uy = u sinθ∴ In horizontal (x) direction
d = uxt + 1/2 gt2 = u cosθ
=
(sinθ cos θ + sin2θ)We want to maximize f(θ) = cosθ sinθ + sin2θ
⇒ f ′θ = − sin2θ + cos2θ + 2 sinθ cosθ = 0
⇒ cos2θ + sin2θ = 0
⇒ tan2θ = −1
or 2θ =
or θ = = 67.5oAlternate :
As shown in figure, the net acceleration of projectile makes on angle 45o with horizontal. For maximum range on horizontal plane, the angle of projection should be along angle bisector of horizontal and opposite direction of net acceleration of projectile.
∴ θ =
= 67.5o Create a free account to view solution
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