A cart is moved horizontally with a constant velocity of 4 m/sec. A ball is thrown from it with a velocity of 4 m/sec and at an angle $θ$ with the horizontal with respect to the cart. Assume the height of the cart is very small, so that the motion of the ball is assumed to be a ground-to-ground projectile. Horizontal range of the ball with respect to the ground is R₁ and that with respect to the cart is R₂. Then

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$R_{1} will be maximum for θ = 60^{\circ}$

$R_{1} will be maximum for θ = 90^{\circ}$

$R_{2} will be maximum for θ = 60^{\circ}$

$R_{2} will be maximum for θ = 45^{\circ}$

answer is A, D.

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Detailed Solution

Range with respect to ground

$\begin{array}{l} R = \frac{2 u_{x} u_{y}}{g} = \frac{2 (4 + 4 \cos θ) (4 \sin θ)}{g} \\ R = \frac{32 \sin θ (1 + \cos θ)}{g} = \frac{32}{g} (\sin θ + \sin θcos θ) \\ R_{max} when \sin θ + \sin θcos θ will be max \\ \Rightarrow \frac{d}{dθ} (\sin θ + \sin θcos θ) = 0 \\ \Rightarrow \cos θ + 2 \cos^{2} θ - 1 = 0 \\ \Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = 60^{\circ} \end{array}$