1
$\begingroup$

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...

$\endgroup$
  • 2
    $\begingroup$ Are you sure part a) is 31 m/s? When I plug in I get 30.03 m/s. $\endgroup$ – Michael M Jan 24 '15 at 18:05
  • $\begingroup$ @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) $\endgroup$ – secondubly Jan 24 '15 at 18:12
  • 1
    $\begingroup$ Shouldn't this be better suited to Physics Stack Exchange? $\endgroup$ – AvZ Jan 24 '15 at 18:22
  • 1
    $\begingroup$ @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. $\endgroup$ – Vladimir Vargas Jan 24 '15 at 18:33
  • $\begingroup$ Well, OP, the formula for time of flight is $\sqrt{\frac{2H}{g}}$. So the answer here will be $\sqrt{\frac{2\times 46}{9.8}}=3.06\ldots$ $\endgroup$ – AvZ Jan 24 '15 at 18:39
2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ OP did use the factor of two, but didn't type that part of the derivation in a way that made it clear (so I made a small edit). The problem appears to be too much "round-off": as you found, the initial velocity is closer to 30 than 31 m/s. So the travel time is closer to 6.1 than 6.3 seconds. (It's a bit hard to see why the homework system accepted OP's velocity, but not their final answer, since both are off by about the same proportion -- 3.2% for the velocity, 2.9% error for the time. Maybe the system is rejecting on absolute errors. ) $\endgroup$ – colormegone Jan 24 '15 at 19:35

protected by user296602 May 18 '18 at 1:39

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.