anonymous
  • anonymous
Ball is kicked from a height of 24 feet at an initial velocity of 16 ft/sec. Approximately how long does it take before the ball lands on the ground?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
So far ive gotten to 0=4t^2 + 4t + but i dunno if it can factor out. Did I screw up somewhere?
whpalmer4
  • whpalmer4
Unspecified is the direction in which the ball is kicked, but that is vital to answering the question...
anonymous
  • anonymous
What??

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More answers

whpalmer4
  • whpalmer4
Without knowing the direction the ball gets that velocity of 16 ft/s, it is impossible to solve the problem.
anonymous
  • anonymous
If you use the Projectile Motion Equation then it is just a matter of plugging in numbers
whpalmer4
  • whpalmer4
The answer is very different if the ball goes straight up, or if it is kicked horizontally, or down toward the ground. No, it's not, unless you know the direction.
anonymous
  • anonymous
Ok, then lets just say it is kicked upwards, now what?
anonymous
  • anonymous
h(t) = -1/2gt^2 + vt + h
whpalmer4
  • whpalmer4
Ball going straight up has equation of motion \[h(t) = -16t^2 + 16t +24\] Ball kicked horizontally from a 24 foot high platform would have equation of motion \[h(t) = -16t^2 + 0t + 24\] Okay, we'll do straight up. You're looking to find the time at which it hits the ground, or \(h(t) = 0\), so \[-16t^2 + 16t + 24 = 0\]
whpalmer4
  • whpalmer4
Factor out a 4: \[-4t^2 + 4t + 6 = 0\]
anonymous
  • anonymous
Ok, I just made a mistake in my math, I did same thing but I got -3t^2 + 4t + 8 I just messed up the 24/4
whpalmer4
  • whpalmer4
I'd probably just do the quadratic formula at this point, with \(a = -4, b = 4, c = 6\) \[t = \frac{ -b \pm \sqrt{b^2 -4ac}}{2a}\]
whpalmer4
  • whpalmer4
one of the values will be negative, and we can ignore that one.
anonymous
  • anonymous
So t= 84.667?
whpalmer4
  • whpalmer4
no...want to show me your work? that's off by about a factor of 45 or so
whpalmer4
  • whpalmer4
that'd be amazing hang time, ball would look like it just hung there on a string or something :-)
anonymous
  • anonymous
Calculator is really not working today :( After recalculating I got 13.166
whpalmer4
  • whpalmer4
write out your numbers for me
whpalmer4
  • whpalmer4
what does the equation look like before you punch it into the calculator?
anonymous
  • anonymous
Cant write it out because of square root symbol. Hold on one sec, I will try it again
whpalmer4
  • whpalmer4
just say sqrt(stuff) where you want the square root -b +/- sqrt(b^2 - 4ac) etc
anonymous
  • anonymous
10.5? my last step was -84/-8
whpalmer4
  • whpalmer4
no. I have to go now, so I'll just show you how I do it: \[a = -4, b = 4, c = 6\]\[t = \frac{ -b \pm \sqrt{b^2 -4ac}}{2a}\] \[t = \frac{-4 \pm\sqrt{(4)^2 - 4(-4)(6)}}{2(-4)} = \frac{-4\pm\sqrt{16+96}}{-8}\] you should be able to finish from there
anonymous
  • anonymous
15
anonymous
  • anonymous
lolol omg I hate math sooooo much thx for your help
whpalmer4
  • whpalmer4
no, how do you get that? 16+96 = 112. 112 = 7*16 so sqrt(112) = sqrt(7)*sqrt(16) = 4sqrt(7) answer is \[\frac{-4\pm4\sqrt{7}}{-8}\]
anonymous
  • anonymous
Ohhh yea the sgrt. I forgot about it
anonymous
  • anonymous
ok, that makes a lot more sense
anonymous
  • anonymous
Would that also be the maximum height that the ball can reach?
whpalmer4
  • whpalmer4
No, the maximum height will be at the vertex of the parabola. If you have a parabola written in the form \[y = ax^2 + bx + c\]the vertex x-coordinate is given by \[x = -\frac{b}{2a}\]and the vertex y-coordinate can be found by plugging that value of \(x\) back into the equation. Also, the vertex is exactly at the midpoint on the x-axis between the two solutions where the parabola = 0. Note that this is NOT necessarily half of the flight time! If you start at a different elevation than you land, the vertex will be before the halfway mark in the flight.

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