if it takes 7 seconds to reach the ground, then its speed at impact is __m/s. the total distance fallen is __m, and its acceleration of fall just before impact is __m/s.

Question

anonymous · Answer

The two equations you use is v=gt and d = 1/2gt2

stormfire1 · Answer

(excluding drag)

For speed at impact:
$$V{f}=V{i}+at$$$$Vf = -9.8m/s^2*7s = -68.6 m/s$$ 
For distance:
$$Yf = Yi + \frac{1}{2}g*t$$$$Yf = 0m - \frac{9.8m/s^2}{2}(7s)=-34.3m$$It's acceleration is constant throughout so it's just g: $$-9.8m/s^2$$

stormfire1 · Answer

I left out the fact that Vi = 0m/s in the first part...since it's assumed something was dropped from 0 velocity

stormfire1 · Answer

If it actually started out with a velocity, you could just plug it in there.

anonymous · Answer

So would the first blank be 70m/s, 350m, and 7m/s2

anonymous · Answer

is that correct

stormfire1 · Answer

Without rounding the blanks would be (in order) -68.6m/s, -34.3m and -9.8m/s^2...they're all negative based on me choosing the y reference being 0 from where the object started to fall.  Hopefully that makes sense