(Solved) Math Help: How long will the ball be in the air.

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A ball is thrown starting at 5.5 feet with a velocity of 35 ft/sec. How long will the ball be in the air before it hits the ground?

My son has gotten this far:

0 = -16t^2+35t+5.5

is the answer .293 seconds? Can you prove it and show your work? It has drained us both terribly. :crazy:
 
I remember having to do loads of those problems for physics last year. Lemme have a look in the textbook.

edit: thrown vertically upwards?
 
Last edited:
Angle is not given, nor is the density of the ball.
 
The proof is the graph. I'm assuming angle is determined by the function-- kind of a dumb thing to leave out. The x-intercept is 2.334733 according to my iPhone. So, the ball went 2.334733 ft. And it took 0.0667 seconds to do that.
 
Yeah, doesn't say..., I pictured a baseball being thrown horizontally.
 
Been a while since I've done this stuff, so I may be a bit rusty.

Assuming thrown vertically upwards:

v(f) = v(i) - at <- v as function of time with constant a

0 = 35ft/sec - 32ft/sec.t
t = 1.09s (time to reach max height; not adjusted for +5.5ft elevation)

Using launch point y(i) as 0 elevation, y(f) = y(i) + v(i).t + 0.5at^2

Max height of ball = y(f) = 0 + 35x1.09 + 16x(1.09^2)
= 38.28 + 16x1.1881 = 57.28 ft + 5.5ft = 62.8 ft

Time to drop 63 feet = 16x(t^2)
ie 62.8 = -0.5at^2
t^2 = (2 x 62.8)/a
t^2 = 125.6/32 = 3.925
t = root 3.9 = 1.98s

Total time in air = 1.98 + 1.09 = 2.97s

Which is ~10x what you have, lol (mind you, being a metric country we use 9.81ms^2, so using ft/s was horrible :lol: )

Not sure if my final answer is the right magnitude, after all 3s of flight seems like quite a lot. =/
 
Nevermind my answer. I totally didn't take hangtime into account. :lol: (and that was the whole point of the question!) :dunce: In what math class is your son?
 
How about just solving for (t)? Solving with roots are messing me up. (I've never used roots the entire time I have been away from school..., my math teacher lied). :D
 
Quadratic equation thingy (-b +- root etc) might do it (disregading the negative result), not 100% on that though.
 
Yeah, tried the quadratic equation and came up with two negatives every time we tired it. UGH....
 
Huh, Omnis might be right. Found this:
http://www.1728.com/quadratc.htm

And put in A, B, and C of -16, 35, and 5.5 and got 2.3347 which would be (t) or time where the ball is at a height of zero along the x axis.
 
Pako
Huh, Omnis might be right. Found this:
http://www.1728.com/quadratc.htm

And put in A, B, and C of -16, 35, and 5.5 and got 2.3347 which would be (t) or time where the ball is at a height of zero.
Click to expand...

Yeah, I simply graphed the function. But I don't know and didn't use the formula or whatever in order to find out how long the ball was in the air with hangtime and all that.
 
It's for maths not physics, so I imagine Omnis' way is probably right.

Where's Indigoman when you need him? :lol:
 
:lol: I hear ya. Well his final is tomorrow and I think it was bugging me more than him, but I will stick with the quadratic equation for my suggestion to him. I think we were messing up the (-) in the equation some where.

Thanks guys.

Oh, this is for his Algebra class.
 
btw: here's what the factorization looks like:
http://www93.wolframalpha.com/input/?i=factor+0%3D+-16t^2%2B35t%2B5.5
 
Well, we just figured out what we were doing wrong last night. We weren't working out the equation correctly. In fact, we were doing every quadratic equation wrong.

This is how it should be solved:
{ -35 ± Sq Root ( 1.2250e+3 + 352 ) } / -32

This is how we were doing it wrong:
-35 ± {Sq Root ( 1.2250e+3 + 352 ) / -32}

I can't believe how much time we spent on that sucker.... Thanks guys for the help. He is now armed with the Quadratic equation for his test tomorrow. :D
 
:lol: That quadratic equation is a bitch if you have to dust it off after many years!
 

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