3.5 Free fall  (Page 4/7)

Solution

1. From [link] , $v^2=v_0^2-2g(y-y_0)$ . With $v=0\text{and}{y}_0=0$ , we can solve for y :
   
   $y=\frac{v_0^2}{\mathrm{-2}g}=\frac{(2.0\times {10}^2\text{m}\text{/}{\text{s})}^2}{\mathrm{-2}(9.8\text{m}\text{/}{\text{s}}^2)}=2040.8\text{m}\text{.}$
   
   This solution gives the maximum height of the booster in our coordinate system, which has its origin at the point of release, so the maximum height of the booster is roughly 7.0 km.
2. An altitude of 6.0 km corresponds to $y=1.0\times {10}^3\text{m}$ in the coordinate system we are using. The other initial conditions are $y_0=0,\text{and}{v}_0=200.0\text{m/s}$ .
   We have, from [link] ,
   
   $v^2={(200.0\text{m}\text{/}\text{s})}^2-2(9.8\text{m}\text{/}{\text{s}}^2)(1.0\times {10}^3\text{m})\Rightarrow v=\pm 142.8\text{m}\text{/}\text{s}.$

Significance

Summary

• An object in free fall experiences constant acceleration if air resistance is negligible.
• On Earth, all free-falling objects have an acceleration g due to gravity, which averages $g=9.81{\text{m/s}}^2$ .
• For objects in free fall, the upward direction is normally taken as positive for displacement, velocity, and acceleration.

Conceptual questions