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I'm given a problem that says a rock hammer was thrown at an upward initial velocity of 4.8m/s and had an initial height of 1.8m.

The acceleration (on the moon) is 1.8m/s^2.

What is the maximum height the hammer will reach?

My initial thought is to find:

s(t) (location)

v(t) (velocity/s'(t))

a(t) (acceleration/v'(t)/s''(t))

The problem is with the information I've got:

s(0) = 1.8m

v(t) = 4.8 m/s (or 4.8/t ??)

a(t) = -1.6 m/s^2 (or -1.6/t^2 ??)

I'm not sure how to approach the problem. Any help is appreciated.

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  • $\begingroup$ I think you mean $v(0)=4.8$. $\endgroup$
    – user122283
    Commented Apr 13, 2014 at 18:09
  • $\begingroup$ Yeah, still I don't know how to get to any equations that might help. $\endgroup$
    – Ethan
    Commented Apr 13, 2014 at 18:14
  • $\begingroup$ Hint: For reasonable $t$, the velocity at time $t$ is $4.8-1.8t$. $\endgroup$ Commented Apr 13, 2014 at 18:14

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Use the basic equation of motion (constant acceleration): $$y-y_0=v(0)t+\dfrac{a(t)t^2}{2}\\ \text{ Given: }a(t)=-1.8,v(0)=4.8,y_0=1.8\\ \implies y=1.8+4.8t-\dfrac{1.8t^2}{2}\\ \implies y^\prime=0+4.8-1.8t\\ \text{When set to $0$, gives max: }y^\prime=0\iff 1.8t=4.8\iff t_{\text{max}}=2\dfrac{2}{3}\\ \implies y_{\text{max}}=1.8+4.8t_{\text{max}}-\dfrac{1.8t_{\text{max}}^2}{2}=1.8+12.8-6.4=8.2m$$ DERIVATION OF EQUATION OF MOTION (constant acceleration): $$v=v_0+at\\ \implies y-y_0=\int vdt=\int(v_0+at)dt=\\ v_0t+\dfrac{a(t)t^2}{2}\text{ (by the fundamental theorem of calculus)}\\ \implies y-y_0=v_0t+\dfrac{a(t)t^2}{2}$$

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  • $\begingroup$ This is great, it helps a lot, thank you. :) I'm on the right track. :D $\endgroup$
    – Ethan
    Commented Apr 13, 2014 at 18:23
  • $\begingroup$ @Ethan Sure!${}{}$ $\endgroup$
    – user122283
    Commented Apr 13, 2014 at 18:23
  • $\begingroup$ @AndréNicolas Yes, I see it. I'll edit it. Thanks! $\endgroup$
    – user122283
    Commented Apr 13, 2014 at 18:43
  • $\begingroup$ @AndréNicolas See my edit. $\endgroup$
    – user122283
    Commented Apr 13, 2014 at 18:44
  • $\begingroup$ You are welcome. $\endgroup$ Commented Apr 13, 2014 at 18:50

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