# Math Help - think its a quadratic equation but not sure hence why i need help

1. ## think its a quadratic equation but not sure hence why i need help

A ball is thrown down at 72 km h–1 speed from the top of a building. The
building is 125 metres tall. The distance travelled before it reach the
ground is as follows,

s = u0t + 1/2gtsquared

where:
u0 = initial velocity (ms -1)
g = acceleration due to gravity (10 ms -2)
t = time (s)

Find the time for the ball to drop to fifth of the height of the building.
&
Find the time for the ball to reach the ground.

2. ## Re: think its a quadratic equation but not sure hence why i need help

Originally Posted by gordonmckee
A ball is thrown down at 72 km h–1 speed from the top of a building. The
building is 125 metres tall. The distance travelled before it reach the
ground is as follows,

s = u0t + 1/2gtsquared

where:
u0 = initial velocity (ms -1)
g = acceleration due to gravity (10 ms -2)
t = time (s)

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

$\color{red}{100 = 72t + 5t^2}$

&

Find the time for the ball to reach the ground.

$\color{red}{125 = 72t + 5t^2}$
... solve each equation for t

3. ## Re: think its a quadratic equation but not sure hence why i need help

need a little bit more help with that ...sorry

4. ## Re: think its a quadratic equation but not sure hence why i need help

Originally Posted by gordonmckee
A ball is thrown down at 72 km h–1 speed from the top of a building. The
building is 125 metres tall. The distance travelled before it reach the
ground is as follows,

s = u0t + 1/2gtsquared

where:
u0 = initial velocity (ms -1)
g = acceleration due to gravity (10 ms -2)
t = time (s)

Find the time for the ball to drop to fifth of the height of the building.
&
Find the time for the ball to reach the ground.
1. Convert $u_0 = 72\ \frac{km}h = 20\ \frac ms$

2. $\frac15$ of the height is 25 m; thus the ball has travelled 100 m. Therefore you have to solve for t:

$100 = 20t + \frac12 \cdot 10\ \frac m{s^2} \cdot t^2$

3. When the ball hits the ground it has travelled 125 m; thus you have to solve for t:

$125 = 20t + \frac12 \cdot 10\ \frac m{s^2} \cdot t^2$

5. ## Re: think its a quadratic equation but not sure hence why i need help

thanks but its the solving part im having trouble with

6. ## Re: think its a quadratic equation but not sure hence why i need help

Originally Posted by gordonmckee
thanks but its the solving part im having trouble with

$ax^2+bx+c=0$

has the solution:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Re-write your equation into the form above. Then plug in the values you know.

Keep in mind that a negative time is not very plausible.

7. ## Re: think its a quadratic equation but not sure hence why i need help

tried it and got 2.899 or -6.899.....am I doing this right or gone wrong somewhere?

8. ## Re: think its a quadratic equation but not sure hence why i need help

That's not at all what I get. Could you show your calculations?

(Of course, the quadratic equation will have two roots but the ball is not likely to hit the ground before it was thrown upward, is it?)

9. ## Re: think its a quadratic equation but not sure hence why i need help

100 = 20t + 5tsquared
a= 5
b= 20
c= -100

x= -20 +or- the sqrt (20squared -4x5x-100) / 2 x 5
= -20 + or - sqrt 2400 / 10
=-20 + or - 48.99 / 10
=2.899 or -6.899

10. ## Re: think its a quadratic equation but not sure hence why i need help

Originally Posted by HallsofIvy
That's not at all what I get. Could you show your calculations?
Those do appear to be the (approximate) roots of the equation for the first problem, unless I'm missing something (maybe you didn't convert the initial velocity to m/s?). Naturally, we would discard the negative root.