So I got this question: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:
How long after the kick is the football at a height of 15m?
Solution
Substitute 15 for h in the formula:
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.
Either
or
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.
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.
What next? What next indeed. What next is you CAREFULLY read the question! Which I believe is
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)
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