Question

object 1 is launched at an initial speed $v_0$ at an angle $\theta$ above the horizontal and reaches a maximum height of $y_1$. object 2 is launched at an initial speed $2v_0$ at the same angle $\theta$, reaching a maximum height of $y_2$. what is the relationship between $y_1$ and $y_2$?
a $y_2 = y_1$
b $y_2=sqrt{2}y_1$
c $y_2 = 2y_1$
d $y_2 = 4y_1$

Explanation:

Step1: Find vertical - initial velocity formula

The vertical - initial velocity is $v_{0y}=v_0\sin\theta$. At maximum height, the final vertical velocity $v_y = 0$. Using the kinematic equation $v_y^2 - v_{0y}^2=-2gy$, we can solve for the maximum height $y$. Rearranging the equation for $y$, we get $y=\frac{v_{0y}^2}{2g}=\frac{(v_0\sin\theta)^2}{2g}$.

Step2: Calculate $y_1$

For object 1 with initial speed $v_0$, the maximum height $y_1=\frac{(v_0\sin\theta)^2}{2g}$.

Step3: Calculate $y_2$

For object 2 with initial speed $2v_0$, the vertical - initial velocity is $v_{0y2}=(2v_0)\sin\theta$. Then the maximum height $y_2=\frac{((2v_0)\sin\theta)^2}{2g}=\frac{4v_0^2\sin^2\theta}{2g}$.

Step4: Find the ratio of $y_2$ to $y_1$

Dividing $y_2$ by $y_1$: $\frac{y_2}{y_1}=\frac{\frac{4v_0^2\sin^2\theta}{2g}}{\frac{v_0^2\sin^2\theta}{2g}} = 4$. So $y_2 = 4y_1$.

Answer:

D. $y_2 = 4y_1$