# Integral Application Word Problem

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.

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

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): $$v=v_0+at\\ \implies y-y_0=\int vdt=\int(v_0+at)dt=\\ v_0t+\dfrac{a(t)t^2}{2}\text{ (by the fundamental theorem of calculus)}\\ \implies y-y_0=v_0t+\dfrac{a(t)t^2}{2}$$
• @Ethan Sure!${}{}$ – user122283 Apr 13 '14 at 18:23