## Summersnow8 one year ago John stands on the edge of a deck that is 40.0 m above the ground and throws a rock straight up that reaches a height of 15.0 m above the deck. What is the initial speed of the rock in m/s? which i solved correctly and it is 17.1 Assuming the rock in the previous question misses the deck on the way down, how fast (in m/s) will the rock be moving when it hits the ground? @ybarrap @lightgrav

1. Summersnow8

i used the equation $2as = V _{f}^{2} - V{i}^{2}$

2. anonymous

so, use that correct answer in the same statement (equation) as before, but solve it for v_f ... remember that (now) a is in the same direction as s (both downward) so 2as is positive.

3. anonymous

(or, you could start with v_i = 0, but watch the rock fall 55m

4. Summersnow8

so, $2 (-9.8)(-40) = Vf ^{2} - (17.1)^{2}$

5. anonymous

looks right so far

6. Summersnow8

so vf would be 32.8?

7. anonymous

yes, 55m is almost 4x the upward 15m (so v should be almost double)

8. ybarrap

You got it! Another way to look at this - $$V_f^2 = V_i^2 + 2gs$$ Where $$V_i=0$$, the speed of the rock at the highest point. $$s$$ is is the distance the rock is above the deck plus the height the deck is above the ground: 15+40=55. Now you have everything to solve. You've got the right approach either way!

9. Summersnow8

so is 32.8 the answer? i was told to use the equation i listed

10. anonymous

His eq'n is the same as yours, just re-arranged for v_f

11. ybarrap

Yep! $$V_f^2 = V_i^2 + 2gs=0-2\times9.81\times55\\ \implies V_f=32.8~m/s$$

12. ybarrap

*Negative, because rock is going down: $$V_f^2 = V_i^2 + 2gs=0-2\times9.81\times55\\ \implies V_f=\color{red}{-}32.8~m/s$$

13. Summersnow8

..... negative?

14. ybarrap

Just indicates direction, downward. Up is positive

15. anonymous

the same direction as the acceleration.

16. Summersnow8

it says it's wrong

17. anonymous

ooh, they asked for _speed_ (how fast) not velocity :-(

18. ybarrap

speed is magnitude, velocity includes directional information. So speed would just be 32.8 m/s