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A simple task, we throw a ball into the air upwards as vertically as possible and record the height of the ball at release, $(height=1m, time = 0)$, and the time the ball hit the ground. $(height=0, time=2.80s)$.

We are trying to solve the equation of the height/time parabola using transformations between standard, vertex, and factored form.

I am not exactly sure where to begin, any help would be appreciated.

(I guess we need to find out the speed of the ball when it was released, how high the ball went and at what time, (vertex))

N. F. Taussig
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Stack
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1 Answers1

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Set the origin at the point of release, and let the initial velocity be $u$

First use $s=ut+\frac 12 at^2$ with $s=-1,t=2.8,a=-9.8$ to get $u$

Then use $v^2=u^2+2as$ with $v=0$ to get the max height $s$ above the point of projection.

Can you finish?

David Quinn
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  • $-1=2.8u-4.9(2.8)^2$, $u=13.36$, $-u^2=2as$, $s=9.108$ – Stack May 20 '17 at 23:29
  • Then we use $s=0.5(u + v)t$ with $u=13.36m/s$, $v=0m/s$ and $s=9.108m$ to find the time when the ball hit max height? – Stack May 20 '17 at 23:37
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    Using exact values, you get $s=9.1105...$ and then $v=u+at\implies t\simeq 1.36$ – David Quinn May 21 '17 at 10:48
  • So, $y=-4.9x^2+13.36x+1$ – Stack May 21 '17 at 16:37
  • But if we transform that to vertex form we get $y=-4.9(x-1.858)^2+10.1$ contradicting both max height and at $1.36s$ no? – Stack May 21 '17 at 16:42
  • Motion is in a vertical line. There's no parabola here – David Quinn May 21 '17 at 16:57
  • So the height/time parabola wont be $y=4.9x^2+13.36x+1$, but the vertex form of the height/time parabola will be $y=a(x-1.36)^2+9.11$, and solving for $a$ using the $y-intercept$, $y=-4.38(x-1.36)^2+9.11$, and in standard form $y=-4.38x^2 +...$ – Stack May 21 '17 at 17:06