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...
$$46\text{m}=v_0t-(5\text{m/s}^2)t^2\tag{1}$$ $$ 0\text{m/s}=v_0-(10\text{m/s}^2)t \tag{2}$$ 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.