Question

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

Answer 1

i used the equation \[2as = V _{f}^{2} - V{i}^{2}\]

Answer 2

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.

Answer 3

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

Answer 4

so, \[2 (-9.8)(-40) = Vf ^{2} - (17.1)^{2}\]

Answer 5

looks right so far

Answer 6

so vf would be 32.8?

Answer 7

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

Answer 8

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!

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

..... negative?

Answer 14

Just indicates direction, downward. Up is positive

Answer 15

the same direction as the acceleration.

Answer 16

it says it's wrong

Answer 17

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

Answer 18

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