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

Question

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. — Michael M, 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) — secondubly, Jan 24 '15 at 18:12
@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. — Vladimir Vargas, 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
@VladimirVargas Hmmm, that would explain the recent large migration of physics exercises in our direction... — colormegone, Jan 24 '15 at 19:24
@RecklessReckoner, yes, people are much more friendly here... Unfortunately there isn't a Physics Stack Exchange for us who like checking solutions and helping people with specific questions. I mean, there are plenty of questions on electromagnetism, electrodynamics, quantum theory, astrophysics, etc that couldn't be asked here — Vladimir Vargas, Jan 24 '15 at 19:36
Possible duplicate of Can someone explain why acceleration is not negative in this problem? — Deusovi, Nov 20 '15 at 04:40

score 2 · Answer 1 · answered Jan 24 '15 at 18:15

2

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

answered Jan 24 '15 at 18:15

Vladimir Vargas

4,635

1

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. ) – colormegone Jan 24 '15 at 19:35

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

1 Answers1