Question

current attempt in progress
the drawing shows projectile motion at three points along the trajectory. the speeds at the points are v1, v2, and v3. assume there is no air resistance and rank the speeds, largest to smallest. (note that the symbol \>\ means \greater than\.)
o v2 > v1 > v3
o v1 > v3 > v2
o v1 > v2 > v3
o v2 > v3 > v1
o v3 > v2 > v1

Explanation:

Step1: Analyze vertical and horizontal velocities

In projectile - motion without air resistance, the horizontal component of velocity ($v_x$) remains constant throughout the motion, i.e., $v_{x1}=v_{x2}=v_{x3}$. The vertical component of velocity ($v_y$) changes due to gravity. At point 1, the vertical velocity $v_{y1}$ is non - zero and positive (going up). At point 2 (the highest point), the vertical velocity $v_{y2} = 0$. At point 3, the vertical velocity $v_{y3}$ is non - zero and negative (coming down).

Step2: Use the speed formula $v=\sqrt{v_x^{2}+v_y^{2}}$

For point 1: $v_1=\sqrt{v_{x1}^{2}+v_{y1}^{2}}$. For point 2: $v_2=\sqrt{v_{x2}^{2}+v_{y2}^{2}}=v_{x2}$ (since $v_{y2} = 0$). For point 3: $v_3=\sqrt{v_{x3}^{2}+v_{y3}^{2}}$. Since $v_{x1}=v_{x2}=v_{x3}$, and $|v_{y1}| = |v_{y3}|$ (symmetry in vertical motion), we have $v_1=\sqrt{v_{x}^{2}+v_{y1}^{2}}$, $v_2 = v_x$, $v_3=\sqrt{v_{x}^{2}+v_{y3}^{2}}$. So $v_1>v_3>v_2$.

Answer:

$v_1 > v_3 > v_2$