Missile 3

The missile penetrated the embankment to a depth of 1.2 m. What was its speed when it hit the surface of the embankment if the movement of the missile in the ground lasted 0.021 seconds and if the movement was uniformly slowed down?

Final Answer:

v =  114.2857 m/s

Step-by-step explanation:

$$\begin{gathered} s=1.2\text{m} \\ t=0.021\text{s} \\ s=\frac{1}{2}\cdot a\cdot t^2 \\ a=\frac{2\cdot s}{t^2}=\frac{2\cdot 1.2}{0.021^2}\doteq 5442.1769{\text{m/s}}^2 \\ v=a\cdot t=5442.1769\cdot 0.021=114.2857\text{m/s} \end{gathered}$$

Try another example