# Max height of thrown object with 12m/s at 1/5 of height

1. Oct 6, 2013

### brundlfly

1. The problem statement, all variables and given/known data
An object is shot vertically upward and has a speed of 12 m/s when it reaches one fifth of its maximum height.
a. What is the initial speed of the object?
b. What is the maximum height of the object?
2. Relevant equations
Free falling objects

3. The attempt at a solution

I used the formula vf^2 = vi^2 - 2g(yf - yi) and got 144 = 0 + 19.6(1/5 - 0), I solved for 1/5 so I have 144/19.6 = 1/5 and got 1/5 = 7.35 meters. This times 5 yields 36.73 meters. This would be it for part b. But I am stuck on part a where I have to find the speed. Any advise on how to approach this? Thank you.

2. Oct 6, 2013

### voko

You got a negative value (because -2g(yf - yi) is negative), and then you probably thought "oops, the square of velocity can't be negative, so let's change the sign and hope for the best". It does not work that way.

You should have asked yourself "why did I get a negative value where only a positive one is possible? Where did I go wrong"?

3. Oct 6, 2013

### brundlfly

Not really, the reason why I ended up with a positive value instead of a negative is because

144 = 0 - 2(-9.6)(1/5 height) = 144 = 0 -(-19.6)(1/5 height) = 144 = 19.6 (1/5 height). If you don't agree let me know.

4. Oct 7, 2013

### voko

The basic kinematic equation is $v_f^2 = v_i^2 + 2 a(s_f - s_i)$. Adapted to free fall, it is either $v_f^2 = v_i^2 + 2g (y_f - y_i)$, where $g$ is then taken negative, or $v_f^2 = v_i^2 - 2g (y_f - y_i)$, where $g$ is then taken positive. But you cannot have both the equation in the form $v_f^2 = v_i^2 - 2g (y_f - y_i)$ and $g$ negative in it.

5. Oct 7, 2013

### nasu

yf is not 1/5.
When the speed is zero, the object has reached the maximum height. (h_max).
When the speed is 12 m/s the object height is 1/5 from this maximum height.

1/5 is not a height in meters (or other unit of length) but a fraction.
Saying that 1/5=7.35 m is nonsense.