anonymous
  • anonymous
A ball is thrown upwards with a speed of 5m/s from a top 20m high building. a. how long does it take the ball to hit the ground? b how fast is the ball moving?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
a. you need to find time so find a kinematic equation that involves distance, gravitational acceleration, and initial velocity but is missing time b. you need to find the final velocity so find a kinematic equation that is missing final velocity but includes the aforementioned data
oOKawaiiOo
  • oOKawaiiOo
d= -1/2at^2 v(final)= V(inital)-at use these two equations......
anonymous
  • anonymous
ok thanks

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anonymous
  • anonymous
i got a weird number
anonymous
  • anonymous
did you calculate it ? what did ypu get
theEric
  • theEric
@oOKawaiiOo , \(a=\dfrac{\Delta v}{\Delta t}=\dfrac{v-v_0}{\Delta t}\\\text{(multiply both sides by }t\text{ to get:)}\\\implies a\ \Delta t=v-v_0\\\text{(add }v_0\text{ to both sides to get:)}\\\implies a\ \Delta t+v_0=v\) Also I believe that your \(d=-\frac{1}{2}at^2\) should come from \(d=d_0+v_0\ t+\frac{1}{2}a\ t^2\). And I would agree that you need to find a good equation. I mean, you need an equation that has what you want (time to hit the ground) and that has all the other variables being something you know! You know \(a=g\) in free fall (only gravitational force), \(v_0=5\ [m/s]\), \(d_0=20\ [m]\), and you want \(\Delta t\) (the time interval). You could say \(t_0=0\) and you want \(t\), as an alternative view. Do you know the equation that is \(v^2=v_0^2+2\ a\ d\)? :) That will get you part B. Then you'll have \(v\) and you can use the definition of acceleration, \(a=\dfrac{\Delta v}{\Delta{t}}\), and solve for \(\Delta t\) for part A. Unless I can think of an equation to get right to part A!`
anonymous
  • anonymous
i had 4 seconds is that right for part a
theEric
  • theEric
Sounds reasonable. I haven't calculated it yet.
theEric
  • theEric
It sounds reasonable. It's not what I got, but I could still be wrong. What did you try?
theEric
  • theEric
Wait, I know I made a mistake!
theEric
  • theEric
I still have a different result...
anonymous
  • anonymous
how did you plug it in
anonymous
  • anonymous
may be im doing something wrong
theEric
  • theEric
I did it backwards, but I'll combine the equations... \(v=-\left|\sqrt{v_0^2+2\ a\ \Delta d}~\right|\) \(a=\dfrac{v-v_0}{t}\implies v=a\ \Delta t+v_0\) Substituting \(v=a\ \Delta t+v_0\) into the first equation, \(v=a\ \Delta t+v_0=-\left|\sqrt{v_0^2+2\ a\ \Delta d}~\right|\\\ \\\ \\\implies\Delta t=\dfrac{-\left|\sqrt{v_0^2+2\ a\ \Delta d}~\right|-v_0}{a}\)
theEric
  • theEric
There must be an equation that I forget or never knew to make part A easier...
anonymous
  • anonymous
thanks a lot
theEric
  • theEric
You're welcome :)

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