3. a rock is thrown straight upward with an initial velocity of 7.2 m/s in a location where the acceleration…

3. a rock is thrown straight upward with an initial velocity of 7.2 m/s in a location where the acceleration due to gravity has a magnitude of 9.81 m/s². to what height does it rise? v:=7.2 m/s a = 9.1 m/s vf

3. a rock is thrown straight upward with an initial velocity of 7.2 m/s in a location where the acceleration due to gravity has a magnitude of 9.81 m/s². to what height does it rise? v:=7.2 m/s a = 9.1 m/s vf

Answer

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v_f^2=v_i^2 + 2a\Delta y$. At the maximum - height, the final velocity $v_f = 0$. The initial velocity $v_i=7.2$ m/s and the acceleration $a=- 9.81$ m/s² (negative because it acts in the opposite direction of motion).

Step2: Rearrange the equation for $\Delta y$

Starting from $v_f^2=v_i^2 + 2a\Delta y$, we can solve for $\Delta y$. First, substitute $v_f = 0$ into the equation: $0 = v_i^2+2a\Delta y$. Then, we get $\Delta y=-\frac{v_i^2}{2a}$.

Step3: Substitute the values

Substitute $v_i = 7.2$ m/s and $a=-9.81$ m/s² into the formula $\Delta y=-\frac{v_i^2}{2a}$. So, $\Delta y=-\frac{(7.2)^2}{2\times(-9.81)}=\frac{51.84}{19.62}\approx2.64$ m.

Answer:

$2.64$ m