Question

current attempt in progress a ball is thrown straight upward with a velocity \\(\vec{v_0}\\) and in a time \\(t\\) reaches the top of its flight path, which is a displacement \\(\vec{y}\\) above the launch point. with a launch velocity of \\(2\vec{v_0}\\), what would be the time required to reach the top of its flight path and what would be the displacement of the top point above the launch point? \\(\bigcirc\\) \\(4t\\) and \\(2\vec{y}\\) \\(\bigcirc\\) \\(2t\\) and \\(4\vec{y}\\) \\(\bigcirc\\) \\(2t\\) and \\(2\vec{y}\\) \\(\bigcirc\\) \\(4t\\) and \\(4\vec{y}\\) \\(\bigcirc\\) \\(t\\) and \\(2\vec{y}\\)

Explanation:

Step1: Analyze vertical motion (time to top)

For vertical motion, final velocity \( v = v_0 - gt \). At the top, \( v = 0 \), so \( 0 = \vec{v}_0 - gt \Rightarrow t = \frac{\vec{v}_0}{g} \). If initial velocity is \( 2\vec{v}_0 \), new time \( t' = \frac{2\vec{v}_0}{g} = 2t \).

Step2: Analyze displacement (height to top)

Displacement formula: \( y = \vec{v}_0 t - \frac{1}{2}gt^2 \). Substituting \( t = \frac{\vec{v}_0}{g} \), we get \( y = \vec{v}_0(\frac{\vec{v}_0}{g}) - \frac{1}{2}g(\frac{\vec{v}_0}{g})^2 = \frac{\vec{v}_0^2}{2g} \). For \( 2\vec{v}_0 \), displacement \( y' = \frac{(2\vec{v}_0)^2}{2g} = \frac{4\vec{v}_0^2}{2g} = 4y \).

Answer:

2t and \( 4\vec{y} \) (corresponding to the option: 2t and \( 4\vec{y} \))