• Jul 25th 2011, 10:19 AM
David Green
I have been asked to calculate the maximum height a ball can achieve after being lauched, using;

h = - 4.9t^2 + 19.6t - 12.6

I knew straight away I required the time, so I did the following;

t = -b + or - square root b^2 - 4ac / 2a

t = - 19.6 + or - square root 19.6^2 - 4 (- 4.9)(- 12.6)
2 x ( - 4.9)

t = - 19.6 + or - square root 384.16 - 246.96
2 x ( - 4.9)

t = - 19.6 + or - square root 137.20
2 x( - 9.80)

t = - 19.6 + or - 11.71
- 9.80

t = - 7.89
- 9.80

t1 = 0.81

t2 = - 19.6 + or - 11.71
- 9.80

t2 = - 31.31
- 9.80

t2 = 3.20

I then used both routes to conclude the height in metres to determine if there was any significant difference, there wasn't so I will use t2;

h = - 4.9 x 3.20^2 + 19.6 x 3.20 - 12.6

h = - 50.18 + 62.72 - 12.6

h = 0

This is the third answer I have had using the above formula?

I have had first 65m, then 34m and now 0 metres?

I can see I must have tapped a interger in wrong in the calculator or not used it correctly, but can I ask, does the above now look like I have got it right?

Thanks

David(Wondering)
• Jul 25th 2011, 11:19 AM
Siron
There's been asked the maximum height the ball can achieve, you're calculating the time wherefore the h=0 and so when the ball reach the ground.
Use:
$t=\frac{-b}{2a}$

Now you're calculating after how much time the ball will reach his highest point and then you've just to enter this time $t$ in the equation.
• Jul 25th 2011, 11:28 AM
David Green
Quote:

Originally Posted by Siron
There's been asked the maximum height the ball can achieve, you're calculating the time wherefore the h=0 and so when the ball reach the ground.
Use:
$t=\frac{-b}{2a}$

Now you're calculating after how much time the ball will reach his highest point and then you've just to enter this time $t$ in the equation.

Hi Siron, so are you saying that the height above the cliff is 7m using the equation you quote?

I see what you mean now, the equation you use is taking the average of the two routes of the t1 and t2 I found, I never thought about averaging the routes?

Thanks

David
• Jul 25th 2011, 11:43 AM
Siron
Yes, the maximum height a ball can achieve is indeed 7m en that's you wanted to calculate. If you want to calculate after how much time the ball will reach the ground and so h=0, then you've indeed to use the quadratic formula.
• Jul 25th 2011, 11:50 AM
David Green
Quote:

Originally Posted by Siron
Yes, the maximum height a ball can achieve is indeed 7m en that's you wanted to calculate. If you want to calculate after how much time the ball will reach the ground and so h=0, then you've indeed to use the quadratic formula.

Thanks for your help, much appreciated.
• Jul 25th 2011, 11:51 AM
Siron
You're welcome! :)
• Jul 25th 2011, 12:40 PM
mathjeet
Or you can use the completing the square method to find the maximum height and time taken to reach maximum height

h = - 4.9t^2 + 19.6t - 12.6

= -4.9 (t^2 - 4t) - 12.6

= -4.9 (t^2 - 4t + 4 - 4 ) - 12.6 Add and subtract square of half the coefficient of linear term to complete the square.

= -4.9(t - 2)^2 + 19.6 - 12.6

h = -4.9(t - 2)^2 + 7

Hence, the maximum height is 7 and after 2 second, which can be found from the vertex (2, 7) of the parabola.
• Jul 25th 2011, 12:51 PM
Siron
True, there are a lot of methods to calculate the maximum/minimum of a parabola. Calculating the first derivative and then $f'(x)=0$ is also a possibilty (but David hasn't learned derivatives yet).