Example 3: When a football is kicked with a vertical speed of 20m/s, its height, h metres, after t seconds is given by the formula:

$\displaystyle h=20t-5t^2$

How long after the kick is the football at a height of 15m?

Solution
Substitute 15 for h in the formula:

$\displaystyle 15=20t-5t^2$

This produces a quadratic equation. The solution of the equation gives the times when the height is 15m.

Collect all the terms on one side of the equation.

$\displaystyle 5t^2-20t+15=0$
$\displaystyle 5(t^2-4t+3)=0$
$\displaystyle t^2-4t+3=0$
$\displaystyle (t-1)(t-3)=0$
Either
$\displaystyle t-1 = 0$
$\displaystyle t=1$
or
$\displaystyle t-3=0$
$\displaystyle t=3$
The football is at a height of 15m twice: first on the way up, 1 s after the kick, then on the way down, 3 s after the kick.
So I got this question:

A baseball is hit with a vertical speed of 31.3m/s. Use the formula in the modelling box following Example 3. How long after being hit is the ball at a height of 25m? Give your answer to 1 decimal place.

How exactly do I form the equation? I got it but I don't think I'm on the right track. Thanks a bunch.

2. Originally Posted by nathan02079
So I got this question:

A baseball is hit with a vertical speed of 31.3m/s. Use the formula in the modelling box following Example 3. How long after being hit is the ball at a height of 25m? Give your answer to 1 decimal place.

How exactly do I form the equation? I got it but I don't think I'm on the right track. Thanks a bunch.
Taking the upwards direction as positive and taking the acceleration due to gravity to be -10 m/s^2 and ignoring things such as air resistance etc. then:

Let an object have an initial vertical speed of u m/s. Then $\displaystyle h = ut - 5t^2$.

3. Thanks so far...
so u is the initial verticle speed
so

$\displaystyle 25=ut-5t^2$
$\displaystyle 5t^2-ut+25=0$

Am I on the right direction here? So what next...
do I use the formula $\displaystyle (a+b)^2=a^2+2ab+b^2$?

4. Originally Posted by nathan02079
Thanks so far...
so u is the initial verticle speed
so

$\displaystyle 25=ut-5t^2$
$\displaystyle 5t^2-ut+25=0$

Am I on the right direction here? So what next...
do I use the formula $\displaystyle (a+b)^2=a^2+2ab+b^2$?
What next? What next indeed. What next is you CAREFULLY read the question! Which I believe is
Originally Posted by nathan02079
So I got this question:

A baseball is hit with a vertical speed of 31.3m/s. Use the formula in the modelling box following Example 3. How long after being hit is the ball at a height of 25m? Give your answer to 1 decimal place.
To make things a little easier I've redded the important bits.

Substitute u = 31.3 (NOT 25). Substitute h = 25. Solve the resulting quadratic equation (it won't factorise in the nice way your example did. I suggest using the quadratic formula or using a graphics calculator)

5. So
$\displaystyle t^2-\frac{ut}{5}+5=0$
$\displaystyle (t^2-2t\frac{u}{10}+\frac{u^2}{10})-\frac{u^2}{10}+5=0$
$\displaystyle (t-\frac{u}{10})^2-\frac{u^2}{10}+5=0$

then...

$\displaystyle t-\frac{u}{10}=\pm\sqrt{\frac{u^2}{10}+5}$

right?

EDIT: Your edit helped me a whole lot...thanks. Solved it now! =)

6. Originally Posted by nathan02079
So
$\displaystyle t^2-\frac{ut}{5}+5=0$
$\displaystyle (t^2-2t\frac{u}{10}+\frac{u^2}{10})-\frac{u^2}{10}+5=0$
$\displaystyle (t-\frac{u}{10})^2-\frac{u^2}{10}+5=0$

then...

$\displaystyle t-\frac{u}{10}=\pm\sqrt{\frac{u^2}{10}+5}$

right?

EDIT: Your edit helped me a whole lot...thanks. Solved it now! =)
Re: edit. Good. And you got two answers, right?

7. Yes, I did. =)

$\displaystyle x=5.5,0.9$

Thank you so much =)

8. When the ball moves upward,

$\displaystyle u=31.3$
$\displaystyle a=g=-10$
$\displaystyle s=25$
$\displaystyle v=?$
$\displaystyle t=?$

$\displaystyle v^2=u^2+2as$
$\displaystyle v^2=(31.3)^2+2(-10)(25)$
$\displaystyle v=21.9$

$\displaystyle s=0.5t(u+v)$
$\displaystyle 25=0.5t[(31.3)+(21.9)]$
$\displaystyle t=0.9 s$

When the ball reaches the max height, $\displaystyle H$,

$\displaystyle u=31.3$
$\displaystyle v=0$
$\displaystyle a=g=-10$
$\displaystyle s=H$

$\displaystyle v^2=u^2+2as$
$\displaystyle 0=(31.3)^2+2(-10)(H)$
$\displaystyle H=48.98$

After the ball reaches the max height and moves downward,

$\displaystyle u=0$
$\displaystyle a=g=10$
$\displaystyle s=48.98-25=23.98$
$\displaystyle t=?$

$\displaystyle s=ut+0.5at^2$
$\displaystyle 23.98=(0)t+0.5(10)t^2$
$\displaystyle t=2.2 s$

Therefore,
$\displaystyle t=0.9 s$ and $\displaystyle t=2.2 s$

If there is any mistake, correct me. Thanks.

9. Originally Posted by nathan02079
Yes, I did. =)

$\displaystyle x=5.5,0.9$

Thank you so much =)

Excuse me, I have a problem.
Your answer is only correct in mathematic steps, but incorrect in actually case right?

10. Originally Posted by SengNee
After the ball reaches the max height and moves downward,

$\displaystyle u=0$
$\displaystyle a=g=10$
$\displaystyle s=48.98-25=23.98$
$\displaystyle t=?$

$\displaystyle s=ut+0.5at^2$
$\displaystyle 23.98=(0)t+0.5(10)t^2$
$\displaystyle t=2.2 s$
We are looking for times when the ball is 25 m above the ground. Not 25 m below the max height.

-Dan

11. Originally Posted by topsquark
We are looking for times when the ball is 25 m above the ground. Not 25 m below the max height.

-Dan

u=0
a=g=10
Displacement of the ball=s=23.98 (Let max h is 0.)
t=?

s=ut+0.5att
23.98=5tt
t=2.2 s

Is it right?

12. Originally Posted by SengNee

u=0
a=g=10
Displacement of the ball=s=23.98 (Let max h is 0.)
t=?

s=ut+0.5att
23.98=5tt
t=2.2 s

Is it right?
As I said before, no, but taking a closer look at what you are doing I realize I was saying so for the wrong reason. My apologies.

For the second time for which the object is at 25 m you have reset your clock to start at 0 s when the ball is at its maximum height. Thus when you find the time it takes for the ball to get back to 25 m, you need to add the time it took for the ball to get to its maximum height.

Do that and you will get the 5.3 s.

-Dan

13. Originally Posted by topsquark
As I said before, no, but taking a closer look at what you are doing I realize I was saying so for the wrong reason. My apologies.

For the second time for which the object is at 25 m you have reset your clock to start at 0 s when the ball is at its maximum height. Thus when you find the time it takes for the ball to get back to 25 m, you need to add the time it took for the ball to get to its maximum height.

Do that and you will get the 5.3 s.

-Dan

I see, thanks a lot.