SOLUTION: Hi. My teacher gave me this problem and i dont understand it: The general formula for the height, h, after t seconds of a projectile launched upward from the ground level with i

Algebra ->  Trigonometry-basics -> SOLUTION: Hi. My teacher gave me this problem and i dont understand it: The general formula for the height, h, after t seconds of a projectile launched upward from the ground level with i      Log On


   

Question 167705: Hi. My teacher gave me this problem and i dont understand it:
The general formula for the height, h, after t seconds of a projectile launched upward from the ground level with initial speed, v is h=vt-1/2gt^2 where g is the gravatational constant (9.8 m/s^2 or 32 ft/s^2). Show that the greatest height of the projectiles is v^2/2g.
thanks.
Found 2 solutions by jim_thompson5910, Earlsdon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
It might be hard to see, but the equation h=vt-%281%2F2%29gt%5E2 can be rearranged to h=-%281%2F2%29gt%5E2%2Bvt and fits the form y=at%5E2%2Bbt%2Bc where a=-%281%2F2%29g, b=v, and c=0


So to find the greatest height, we need to find the vertex.


To do that, we use this formula


t=%28-b%29%2F%282a%29


t=%28-v%29%2F%282%28-%281%2F2%29g%29%29 Plug in a=-%281%2F2%29g, b=v


t=%28-v%29%2F%28-g%29 Multiply and reduce


t=v%2Fg Reduce


So the max height will occur when the time is t=v%2Fg


---------------------------------


h=vt-%281%2F2%29gt%5E2 Go back to the original equation


h=vt-%281%2F2%29g%28v%2Fg%29%5E2 Plug in t=v%2Fg


h=v%28v%2Fg%29-%281%2F2%29g%28v%5E2%29%2F%28g%5E2%29 Square v%2Fg to get v%5E2%2Fg%5E2


h=v%28v%2Fg%29-%281%2F2%29%28v%5E2%29%2F%28g%29 Cancel like terms. (one pair of "g" terms cancel)


h=v%5E2%2Fg-%28v%5E2%29%2F%282g%29 Multiply


h=%282v%5E2%29%2F%282g%29-%28v%5E2%29%2F%282g%29 Multiply the first fraction by 2%2F2


h=%282v%5E2-v%5E2%29%2F%282g%29 Combine the fractions


h=%28v%5E2%29%2F%282g%29 Subtract


So at the time t=v%2Fg the max height will be h=%28v%5E2%29%2F%282g%29

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Prove that the maximum height of the projectile launched upward is: v%5B0%5D%5E2%2F2g.
Starting with the function for the height (as a function of time, t) of a projectile launched upward from an initial height of h%5B0%5D with an initial velocity of v%5B0%5D:
h%28t%29+=+-%281%2F2%29gt%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D
In the given problem:
h%5B0%5D+=+0 Projectile is launched from the ground, so you have:
h%28t%29+=+-%281%2F2%29gt%5E2%2Bv%5B0%5Dt
When graphed, this equation describes a parabola that opens downward (negative coefficient of t%5E2 so the vertex is a maximum.
The abscissa (corresponds to the x-coordinate) of the vertex is given by:
x+=+-b%2F2a where the a and b come from the general form of a quadratic equation:f%28x%29+=+ax%5E2%2Bbx%2Bc, but, in this problem, the independent variable is t rather than x, and a+=+-%281%2F2%29g and b+=+v%5B0%5D, so you have:
t+=+-v%5B0%5D%2F2%28-%281%2F2%29g%29 This is the time at which the projectile will be at its maximum height. Let's call this t%5Bm%5D. Simplifying this, you get:
t+=+v%5B0%5D%2Fg Now substitute this value of t into the original equation h%28t%29+=+-%281%2F2%29gt%5E2%2Bv%5B0%5Dt to find the maximum height of the projectile:
h%28t%5Bm%5D%29+=+-%281%2F2%29g%28v%5B0%5D%2Fg%29%5E2%2Bv%5B0%5D%28v%5B0%5D%2Fg%29 Simplifying:
h%28t%5Bm%5D%29+=+-%281%2F2%29g%28v%5B0%5D%5E2%2Fg%5E2%29%2Bv%5B0%5D%5E2%2Fg
h%28t%5Bm%5D%29+=+-v%5B0%5D%5E2%2F2g+%2B+v%5B0%5D%5E2%2Fg
h%28t%5Bm%5D%29+=+v%5B0%5D%5E2%2Fg QED
The maximum height h%5Bm%5D attained by the projectile in this problem is:
highlight%28h%5Bm%5D+=+v%5B0%5D%5E2%2F2g%29