-1
$\begingroup$

A baseball is given an initial velocity of $34 m/s$. It attains its maximum height after $5.5 s$. What is the maximum height that it reaches?

This question was on a high school physics test.

It seems to me, we could just work out the distance from

$d = \frac{1}{2}at^2 = \frac{1}{2}(9.8)(5.5)^2 = 148.225$,

(sig figs notwithstanding).

However, if I want to work out the vertical component of the velocity:

$v= \sqrt{2ad} = \sqrt{2(9.)(148.225)}= 53.9$

The two parameters, $t=5.5$ and $v_0=34$ appear to be at odds with each other, because we are not given the angle of the trajectory, and the vertical component could not be $53.9$ given this $v_0$.

So, am I having a bad hair day, or is there something askew with this question?

Edit: Thank you for the responses below. I came across this question while tutoring a student in Physics. I wanted to run it by the community here before telling him that it's an error. The only way the question parameters seem to work is if we drop the assumption that $g=9.80 m/s^2$ -- a pretty bold move. Also, we have to make an apparently baseless assumption about the angle of the trajectory.

Improve this question
$\endgroup$
4
  • $\begingroup$ since $y=y_0+v_0t+\frac{1}{2}at^2$ and $v_0\ne 0$ in your problem your calculation of $d$ is not correct. $\endgroup$
    – ZeroTheHero
    Commented Apr 4, 2017 at 2:13
  • 2
    $\begingroup$ @ZeroTheHero For projectile motion time up equals time down when returning to the same vertical position, so you can start at the top with a zero velocity and fall 5.5 s. $\endgroup$
    – Bill N
    Commented Apr 4, 2017 at 2:20
  • 1
    $\begingroup$ @BillN You are right and it's clearly too late for me comment on such things :) $\endgroup$
    – ZeroTheHero
    Commented Apr 4, 2017 at 4:04
  • $\begingroup$ Looking over this again after $t \approx 2 yrs$... This question was closed because it looks like a homework question with little effort? Puh-leeeze! $\endgroup$
    – MathAdam
    Commented Jan 23, 2019 at 4:33

1 Answer 1

Reset to default
1
$\begingroup$

You are right - the question was written poorly. Even if the ball is thrown vertically, it would take less than 3.5 seconds to reach the height of its arc on Earth.

However, while we usually assume physics problems take place with $g\approx 10 m/s^2$, if we drop that assumption and assume the ball is thrown vertically and ignore any drag or forces other than gravity, we can answer the question. You should find a height of 93.5 meters in that case.

Improve this answer
$\endgroup$
2
  • $\begingroup$ You seem to be suggesting that we drop the "assumption" that $g \approx 10m/s$. That seems rather an arbitrary assumption, solely for the purpose of avoiding the more obvious conclusion, that the question is faulty. $\endgroup$
    – sammy gerbil
    Commented Apr 4, 2017 at 5:30
  • $\begingroup$ This answer works, if we allow for a gravitational acceleration of $6.18 m/s^2$. I'm not aware of any planet in this solar system that satisfies this condition. Perhaps the baseball game is being played on an accelerating spacecraft. I should also add that the original question said that the player had hit a home run. Did she hit the ball straight up... and run around the diamond in 11 seconds? Further evidence that this game did not take place on earth. Thank you Mark for affirming me in my suspicion that the question is dodgy. Yes, we could have left it there, but where's the fun in that? $\endgroup$
    – MathAdam
    Commented Apr 4, 2017 at 5:43

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