Calculate how long a ball will be in the air after being thrown

So I'm doing some online homework, and have done this specific problem 3 different times and gotten the same answer, but the answer I get seems to be wrong? The problem is as follows:

(a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 46 m? (b) How long will it be in the air?

Now part a I solved, and got 31 m/s (which is correct)

Using that result for part b, I did the following:

(b) \begin{align} t & = \frac{v-v_{0}}{a} \\ & = \frac{0-31}{-9.8} \\ & = 3.16326530... \\ \text{total travel time} \ & = (3.16326530...) \cdot 2 \\ & = 6.3 s \end{align}

Apparently this answer is incorrect? I'm not sure why though...

• Are you sure part a) is 31 m/s? When I plug in I get 30.03 m/s. Jan 24 '15 at 18:05
• @MichaelM. Well I probably rounded something in the intermediate step that I shouldn't have - I'll try using the actual number and see what I get (strangely enough, the computer accepted 31 m/s) Jan 24 '15 at 18:12
• Shouldn't this be better suited to Physics Stack Exchange?
– AvZ
Jan 24 '15 at 18:22
• @AvZ they will close the question if it is asked at PSE. They adopted some rules that are not very friendly with people with doubts about specific problems or homework exercises. Jan 24 '15 at 18:33
• Well, OP, the formula for time of flight is $\sqrt{\frac{2H}{g}}$. So the answer here will be $\sqrt{\frac{2\times 46}{9.8}}=3.06\ldots$
– AvZ
Jan 24 '15 at 18:39

$$46\text{m}=v_0t-(5\text{m/s}^2)t^2\tag{1}$$ $$0\text{m/s}=v_0-(10\text{m/s}^2)t \tag{2}$$ from the equations of motion.
Now from $(2)$ we get $$v_0=(10\text{m/s}^2)t.$$ Then $$46\text{m}=(10\text{m/s}^2)t^2-(5\text{m/s}^2)t^2=(5\text{m/s}^2)t^2\Longrightarrow t=+\sqrt{\frac{46}{5}}\text{s}\approx 3\text{s}.$$ Therefore $v_0\approx 30\text{m/s}$.
I used $g=-10$m/s$^2$
The time in the air is twice the one I found solving for $t$ where the height is maximum.