a rocket accelerates at 15 m/s² from rest for 21 m on a frictionless horizontal surface. the rocket stops…

a rocket accelerates at 15 m/s² from rest for 21 m on a frictionless horizontal surface. the rocket stops firing at the cliff and falls freely from a height of 45 m. a = 15 m/s²

a rocket accelerates at 15 m/s² from rest for 21 m on a frictionless horizontal surface. the rocket stops firing at the cliff and falls freely from a height of 45 m. a = 15 m/s²

Answer

Explanation:

Step1: Find the velocity when the rocket stops firing

Use the equation $v^{2}=v_{0}^{2}+2ax$. The rocket starts from rest ($v_{0} = 0$), $a = 15\ m/s^{2}$ and $x=21\ m$. $v=\sqrt{2ax}=\sqrt{2\times15\times21}=\sqrt{630}\approx 25.1\ m/s$

Step2: Analyze the free - fall motion

After the rocket stops firing, it is in free - fall with an acceleration $g = 9.8\ m/s^{2}$ and falls from a height $h = 45\ m$. Use the equation $v_{f}^{2}=v^{2}+2gh$, where $v$ is the velocity at the end of the acceleration phase and $h$ is the height of free - fall. $v_{f}=\sqrt{v^{2}+2gh}=\sqrt{630 + 2\times9.8\times45}=\sqrt{630+882}=\sqrt{1512}\approx38.9\ m/s$

Answer:

The velocity of the rocket just before it hits the water is approximately $38.9\ m/s$