The Observation Deck

I had a problem on my homework that I just can't get. Asked my brother (engineering major) to explain it to me but he too noticed that you can't have two unknown variables in one equation, and solve for one
.
I've tried everything from using quadratic equations to the kinematic equations that I learned in physics last year. there's still a number missing!

anyway here's the problem:

The height of a projectile fired upward is given by the formula
S=V0t-16t^2
where
s=height
V0 is the initial velocity
and t=time
Find the time for a projectile to return to earth if it has an initial velocity of 200 ft/s.

if you use the kinematic equation of Vf^2=V0^2+2*a*d then you can figure out Vf, which is the final velocity of the object.

first you convert everything to metric units.
If Vf^2 is zero, and V0 (initial velocity) is 60.96 m/s. (200 ft/s= 60.96 m/s)
acceleration would be -9.8 m/s because gravity is acting against the projectile, therefore being negative. and that also is squared.

you will come up with a number for distance which is 189.6 meters.

then you use the distance equation:

d=(V0+Vf/2)*t
plug in the numbers
and come up with time = 6.22 seconds, and that's to get to the top of the arch. now it's gotta come down.
so you use the final velocity equation again.

Vf^2= 0+2*(9.8m/s)*189.6

Vf^2=3716.16
take the square root of both sides and you come up with 60.96

AHHH!!! it's all a circle.

I know the answer but it doesn't help if I don't know how to get it!
The answer is 12.5 seconds to come back to earth after being launched.

ANY ideas anyone? I hope there's some math wizards out there somewhere!

DUDE! Thanks for reminding me why I was a history major. It may never get me a job but at least I earned a college degree without ever having to solve anything like that. Sorry I can't help, hopefully Stephen Hawkins will be checking in to the Deck to lend a hand in solving that equation.

haha, this is normal stuff for me. I'm not in a accelerated class either. It's really simple, but something's missing and it's making me frustrated! I can't figure it out.

this is going to bother me until I visit my physics teacher tomorrow!

S=vt - 16tx t

given v = 200 and S= 0 ie ZERO

0 = 200t - 16 txt

16txt = 200t

16t = 200

t = 12.5

Ivor, this is obviously a trick question. The teacher failed to give you all of the variables needed.

If, say, the projectile was fired indoors, then at 200 ft/s, it would more than likely lodge itself into the ceiling, making the correct answer "never".

If, say the projectile was fired on a path parallel to the earth surface, then regardless of the initial velocity, the time to fall to earth is simply determined by figuring out the initial height from which the shot was fired, and taking into account the gravitational pull on the projectile.

Does the projectile have a parachute attached?

do you need a permit to fire said projectile?

I'm afraid, the best you can do is write in big red letters across your paper "INSUFFICIENT INFORMATION".

If my memory serves me right....

the initial velocity is purely upward motion. the force of gravity is a scalar unit that affects only the upward and downward acceleration of the object.

As the object is fired up with an initial velocity, gravity is a negative acceleration on it. When its velocity equals zero the maximum height of the object is reached. Then it begins accelerating downward from a known height and an initial velocity of zero.

That's what I was going to say

Mr.Kampe,

I asked my old physics teacher today how to do it, and I completely forgot about the quadratic formula.

or your method, it also works.

it was so simple that's what made me mad because I just couldn't figure it out at first!

Ryan you are also correct. in one of the kinematic equations the force (a=accel.) is -9.8m/s. that's equal to the gravitational pull. if you switch it over to feet you get another number.

But this didn't need the big 4 equation, it needed the quadratic equation! or you could set S to equal zero.

High school certainly isn't getting any easier as years go by, eh?

"Background" of why set s =0 (ie if on ground then s=0 )

So S=0 at start and also at end which quadratic eqn yield 0 = 200t-16t*2

then 0 = t ( 200- 16t ) and thus solution are t=0 and t =12.5 .

ie t= 0 ( ie start ) and t = 12.5 ( at end ) for initial velocity of 200

Sometimes in science and for that matter life ( and posting on LO ) - one makes solution or opinion too complicated .

Good Luck

I guess I won't see the answer on Ren and Stimpy anytime soon eh?