An archer shoots an arrow from a height of 4. | vedclass

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(B) Let the initial position be at height $h = 4.2 	ext{ m}$. The initial velocity components are $u_x = v \cos 30^{\circ} = 40 	imes \frac{\sqrt{3}

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$84 \sqrt{3} 	ext{ m}$

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(B) Let the initial position be at height $h = 4.2 	ext{ m}$. The initial velocity components are $u_x = v \cos 30^{\circ} = 40 	imes \frac{\sqrt{3}}{2} = 20\sqrt{3} 	ext{ m/s}$ and $u_y = v \sin 30^{\circ} = 40 	imes \frac{1}{2} = 20 	ext{ m/s}$.Using the equation of motion for the vertical direction,taking the downward direction as positive:$y = u_y t + \frac{1}{2} g t^2$$4.2 = -20t + \frac{1}{2} (10) t^2$$5t^2 - 20t - 4.2 = 0$Multiplying by $10$ to simplify: $50t^2 - 200t - 42 = 0$. This is incorrect; let's use the standard convention where upward is positive:$-4.2 = 20t - \frac{1}{2} (10) t^2$$5t^2 - 20t - 4.2 = 0$$t = \frac{20 \pm \sqrt{(-20)^2 - 4(5)(-4.2)}}{2(5)} = \frac{20 \pm \sqrt{400 + 84}}{10} = \frac{20 \pm \sqrt{484}}{10} = \frac{20 \pm 22}{10}$.Since $t > 0$,$t = \frac{42}{10} = 4.2 	ext{ s}$.The horizontal distance $R$ is given by $R = u_x t = (20\sqrt{3}) 	imes 4.2 = 84\sqrt{3} 	ext{ m}$.

An archer shoots an arrow from a height of $4.2 \text{ m}$ above the ground with a speed of $40 \text{ m/s}$ at an angle of $30^{\circ}$ with the horizontal,as shown in the figure. Determine the total horizontal distance $R$ covered by the arrow when it hits the ground. (Take $g = 10 \text{ m/s}^2$)

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