A small object is thrown at an angle 45^{circ | vedclass

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(D) Let the object be at point $B(x, y)$ after time $t = \sqrt{2} \,s$. The horizontal displacement is $x = u_x 	imes t = (v_0 \cos 45^{\circ}) 	ime

Vedclass · Accepted Answer

$5 \,m/s$

Vedclass · Answer

(D) Let the object be at point $B(x, y)$ after time $t = \sqrt{2} \,s$. The horizontal displacement is $x = u_x 	imes t = (v_0 \cos 45^{\circ}) 	imes \sqrt{2} = v_0 	imes \frac{1}{\sqrt{2}} 	imes \sqrt{2} = v_0$. The vertical displacement is $y = u_y t - \frac{1}{2} g t^2 = (v_0 \sin 45^{\circ}) \sqrt{2} - \frac{1}{2} (10) (\sqrt{2})^2 = v_0 - 10$. The displacement vector $\vec{OB}$ has magnitude $OB = \sqrt{x^2 + y^2} = \sqrt{v_0^2 + (v_0 - 10)^2}$. The average velocity magnitude is given by $v_{	ext{avg}} = \frac{OB}{t} = |v_0|$. Thus, $OB = |v_0| t = v_0 \sqrt{2}$. Squaring both sides: $v_0^2 + (v_0 - 10)^2 = (v_0 \sqrt{2})^2$. $v_0^2 + v_0^2 - 20v_0 + 100 = 2v_0^2$. $2v_0^2 - 20v_0 + 100 = 2v_0^2$. $20v_0 = 100$, which gives $v_0 = 5 \,m/s$.

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