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A ball is thrown down at 72km h-1 speed from the top of a building. The building is 125 metres tall. The distance travelled before it reached the ground is as follows... $$s = U_0 t + \frac15gt^2$$ where

  • Uo = initial velocity (m s-1)
  • g = acceleration due to gravity (10m s-2)
  • t = time(s)

Find the time for the ball to drop to a fifth of the height of the building.

I have worked out the following but I need it checking, if someone could check it I would appreciate it.

1/5 of the height of the building is 25 meters. So the ball has travelled 100m.

using these equations... x=-b(+or-) Squareroot b^2 -4ac/2a


a=5 b=20 c=-100

Uo=20ms g=10ms S=100m t=20

with the above I have got...

x=-20(+or-) sqroot of 20^2 -4x5-100/2x5

I get the following...

x=-20(+or-) sqroot 2400/10

then taking the positive answer I get





20-48.99/10 =-2.899 FOR THE ANSWER.

As the time cannot be negative... the answer is 6.899 secs, but I know I've gone wrong somewhere can anybody clarify this equation for me please?

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Here's a MathJax tutorial :) –  Shaun Jul 10 at 20:14
You already asked this question at math.stackexchange.com/q/858124/18398. Why do you ask it again? –  Joel Reyes Noche Sep 21 at 13:25

2 Answers 2

up vote 2 down vote accepted

All you have to do is solve for $t$ in the equation

$s(t) = U_{0}t + \dfrac{1}{2}gt^{2}$. You found $U_{0} = 20$ m/s. Also, as you correctly noted, the ball will have travelled 100 m. So we get:

$100 = 20t + \dfrac{1}{2}10t^{2}$

$100 = 20t + 5t^{2}$

Making one side $0$ gives:

$0 = -100 + 20t + 5t^{2}$, or rearranged, $5t^{2} + 20t - 100 = 0$.

Using the quadratic formula on this gives $a = 5$, $b = 20$, and $c = -100$, so:

$t = \dfrac{-20 \pm \sqrt{20^{2} - 4*5*(-100)}}{2*5} = \dfrac{-20 \pm \sqrt{2400}}{10}$

And this gives $t = -6.89$ seconds, and $t = 2.89$ seconds, roughly. As you said, we throw out the negative solution. Your problem is, in the end, you ignored the negative in front of your 20. It should be $\dfrac{-20 + 48.99}{10}$, where you had written $\dfrac{20 + 48.99}{10}$. That explains why your $6.899$ is positive and $2.899$ is negative. The answer is $t = 2.899$ seconds.

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The $1/5$ was a typo, $1/2$ was intended and used further down in the problem. –  Ian Jul 10 at 19:34
Thanks for letting me know. I updated my answer. –  MathIsHardNoItsNot Jul 10 at 19:38

I think you should set up your signs in a more sensible way: $g=-10 m/s^2$, $v_0 = -20 m/s$, and $h_0 = +125 m$. Then your equation is

$$-5 t^2 - 20 t + 125 = 25$$

which is equivalent to

$$5t^2 + 20t - 100 = 0$$

This makes you less prone to an error in the setup. You didn't make one, but setting things up the way you did makes one rather likely.

Your mistake was losing track of a minus sign: your solutions should be $\frac{-20 \pm 48.99}{10}$, not $\frac{20 \pm 48.99}{10}$. This gives a positive solution of $t \approx 2.899$, which when plugged back in gives about $0.001$, which is from rounding error.

(You also typo'd $1/5$ instead of $1/2$ at the top.)

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