Rosanne drops a ball from a height of 400 ft. Find the ball's average height and its average velocity between the time it is dropped and the time it strikes the ground.
My trial...
So, I tried to use average value theorem for integrals. I took acceleration as positive $32$ ft m/s$^2$ (as velocity increases). By integrating it, I found velocity as $v=32t + C$ (with $C=0$). I integrated it one more time and got the distance $16 t^2 = 400$. And now $t = 5$ sec. After that, I tried to use average value of the function theorem. Somehow got the wrong answer. For finding H average and I did the following. (1/5)*(integral(from 0 to 5) Htdt. For this integral I found the answer 400/3 while answer given is 800/3
$\endgroup$ 22 Answers
$\begingroup$Assuming $v_0=0$, we have that
$h(t)=400-\frac12gt^2 \implies t_{max}=\sqrt{\frac{800}{g}}=20\sqrt{\frac{2}{g}}$ time to strike the ground
$v(t)=gt \implies v_{max}=20\sqrt{2g} \implies \bar v=\frac{v_0+v_{max}}{2}$
and
$$\bar h = \frac{\int_0^{t_{max}}h(t) dt}{t_{max}}$$
$\endgroup$ 3 $\begingroup$Let $H=400\text{ft}$.$$h(t)=H-\frac12gt^2$$ Average height = particular height $\times$ time during which it was on that height/ total time.$$\frac12gT^2=H\implies T=\sqrt{\frac{2H}g}$$Or mathematically$$<h>=\frac{\int_0^Th(t)dt}{\int_0^Tdt}$$$$<h>=\frac{\int_0^T(H-\frac12gt^2)dt}T=H-\frac{\frac16g\big[t^3\big]_0^T}T=H-\frac{H}3=\frac{2H}3=\frac{800}3\text{ft}$$$v(t)=gt.$So, similarly $$<v>=\frac1T\int_0^Tgtdt=\frac HT=\sqrt{\frac{Hg}2}$$
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