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(D) In projectile motion, the horizontal component of velocity remains constant throughout the flight.Let the initial speed be $u$. The horizontal com

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$20\sqrt{2}$

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(D) In projectile motion, the horizontal component of velocity remains constant throughout the flight.Let the initial speed be $u$. The horizontal component of the initial velocity is $u_x = u \cos 60^{\circ}$.Given that at a certain point, the speed is $v = 20 	ext{ m/s}$ at an angle of $45^{\circ}$ with the horizontal, the horizontal component of this velocity is $v_x = v \cos 45^{\circ} = 20 \cos 45^{\circ} = 20 	imes \frac{1}{\sqrt{2}} = \frac{20}{\sqrt{2}} 	ext{ m/s}$.Since $u_x = v_x$, we have:$u \cos 60^{\circ} = \frac{20}{\sqrt{2}}$$u 	imes \frac{1}{2} = \frac{20}{\sqrt{2}}$$u = \frac{40}{\sqrt{2}} = 20\sqrt{2} 	ext{ m/s}$.

$A$ projectile is thrown upward at an angle $60^{\circ}$ with the horizontal. The speed of the projectile is $20 \text{ m/s}$ when its direction of motion is $45^{\circ}$ with the horizontal. The initial speed of the projectile is . . . . . . in $\text{m/s}$.

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