Answer:
a) t = 2.14 s
b) = -24.972 m/s
Explanation:
We know that the initial height of the package is 31 m, and the final height is 0 m, additionally, its initial velocity is the velocity of the helicopter, so it is -4m/s because it is descending. Finally, the acceleration is due to gravity, so it is -9.8 m/s².
Now, we can use the following equation to find the time to reach the ground
[tex]y_f=y_i+v_it+\frac{1}{2}at^2[/tex]Where y is the height, v is the velocity, a is the acceleration and t is the time. Replacing the values, we get:
[tex]\begin{gathered} 0=31-4t+\frac{1}{2}(-9.8)t^2 \\ 0=31-4t-4.9t^2 \end{gathered}[/tex]So, we can find the roots for the quadratic equation as:
[tex]\begin{gathered} t=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(-4.9)(31)}}{2(-4.9)} \\ t=2.14\text{ and t = -2.96} \end{gathered}[/tex]Since t = -2.96 has no sense in this case, the answer is t = 2.14. It means that the package takes 2.14 seconds to reach the ground.
On the other hand, we can calculate the final velocity using the following equation:
[tex]v_f=v_i+at[/tex]Replacing the values, we get:
[tex]\begin{gathered} v_f=-4+(-9.8)(2.14) \\ v_f=-4-20.972 \\ v_f=-24.972\text{ m/s} \end{gathered}[/tex]So, the velocity just before it lands is -24.972 m/s
