Quadratic Functions

[dmarie8908]

in my school's math classes we are doing the Coach books for the PSSA's (pennsylvania tests). most of the topics have never been covered since it is designed for juniors (i am a sophomore). the problem says: "Two objects are dropped: one from a height of 300ft., the other from a height of 600ft. To the nearest tenth of a second, how much after the first object hits the ground will the second object hit?"


the only thing i can think of trying is converting it. 9.8/s^2 * 1/.3048 . i have never attempted these kind of problems and have only seen them once or twice. is there another way to solve it?

 

[daon]

The distance of a "dropped" object is given by:

\(\displaystyle d = \frac{gt^2}{2} =\)\(\displaystyle 16t^2\) (32 feet/sec² is acceleration due to gravity)

So object 1 is:
\(\displaystyle 300 = 16t^2 \Rightarrow t = \sqrt{\frac{300}{16}}\)

And object 2 is:
\(\displaystyle 600 = 16t^2 \Rightarrow t = \sqrt{\frac{600}{16}}\)


Hope that helps.

edit: I see you are using 9.8 meters/s² so just replace all 32's with 9.8s and 16s with 4.9's.

 

[tkhunny]

Welcome! Fellow Pennsylvanian...

One MUST ask why the Pennsylvania System for School Assesment has become a standard for Student performance, but I digress.

Nice try on the 9.8, but that gives you a bit of a problem, that's 9.8 meter/sec^2 amd you are given height in feet. You'll have to switch something to the same units. Maybe 32 ft/sec^2.

If you drop things, rather than throwing them, it's all a bit more simple. The distance traveled in t seconds is just ½*g*t^2.

Solving the short one: 300 = ½*(32 ft/sec^2)*t^2 ==> t = 4.330 sec
Solving the long one: 600 = ½*(32 ft/sec^2)*t^2 ==> t = 6.124 sec

Since you can use a calculator with a square root button, these should not be tough to solve, if you remember the formula.

Subtraction is the easy part: 6.124 sed - 4.330 sec = 1.794 sec

Good Luck