Question

a rocket is fired upward with an initial velocity of 35 m/s. find the rockets maximum altitude. 1.8 m 63 m 62.5 m it will never leave the earth.

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v^{2}=v_{0}^{2}+2a\Delta y$. At the maximum altitude, the final velocity $v = 0$. The acceleration $a=-g=- 9.8\ m/s^{2}$ (negative because it acts in the opposite direction of motion), and the initial velocity $v_{0}=35\ m/s$.

Step2: Rearrange the equation for $\Delta y$

Starting from $v^{2}=v_{0}^{2}+2a\Delta y$, we can solve for $\Delta y$. Rearranging gives $\Delta y=\frac{v^{2}-v_{0}^{2}}{2a}$.

Step3: Substitute the values

Substitute $v = 0$, $v_{0}=35\ m/s$, and $a=-9.8\ m/s^{2}$ into the formula: $\Delta y=\frac{0 - 35^{2}}{2\times(-9.8)}=\frac{- 1225}{-19.6}=62.5\ m$.

Answer:

C. 62.5 m