kinetic energy approximation

Question

If a pitcher throws a pitch at a velocity $v_0$, then the kinetic energy is $E_0=\frac 12mv_0^2$. If the pitcher releases the pitch from x feet higher, then we will suppose that he can readjust his delivery to hit the same spot over home plate with the same initial speed. But this time, there is added energy from the drop in height x, so the total kinetic energy is

$$\frac 12mv^2=E=E_0+E_1$$. Find the formula for $v$ in terms of $v_0$ and $x$. Then use a linear approximation to estimate the difference $v−v_0$, which is the gain in velocity due to the extra height.
could anyone provide hints? very confused.

Answer 1

Your velocity $v$ has two components, the $x-$ and $y-$direction. Assuming no friction, the $x-$component is constant, while there will be a uniform acceleration $g$ down due to gravity. You can then calculate the $y-$velocity at the moment the pitch reaches the home plate.

You now have two velocity components, but that doesn't mean that $E = E_x + E_y$. Can you see why? This is where the linear approximation comes into play.

Answer 2

We have

$$E=\frac12mv^2=\frac12mv_0^2+mgx$$

from which we can write

$$v=v_0\left(1+\frac{2gx}{v_0^2}\right)^{1/2} \tag 1$$

Assuming that $\frac{2gx}{v_0^2}<<1$, we approximate the square root in $(1)$ as

$$\left(1+\frac{2gx}{v_0^2}\right)^{1/2}\approx. 1+\frac{gx}{v_0^2} \tag 2$$

where the approximation error is of order $\left(\frac{2gx}{v_0^2}\right)^2$. Using $(2)$ in $(1)$ reveals that

$$\bbox[5px,border:2px solid #C0A000]{v-v_0\approx. \frac{gx}{v_0}}$$

which is the linear approximation of the uplift in velocity from the increased elevation $x$.

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A second way to arrive at the same result is to write

$$\frac12m(v^2-v_0^2)=mgx\implies(v-v_0)=\frac{2gx}{v+v_0} \tag 3$$

Assuming that $v\approx. v_0$ when $x$ is "small," we replace $v$ in the denominator of the right-hand side of $(3)$ with $v_0$ and obtain

$$v-v_0\approx.\frac{gx}{v_0}$$

as expected!

Answer 3

Notice, due to drop through a height $x$, with initial velocity $v_0$, the velocity $v$ is increased constantly under earth's gravitational acceleration $g$.

Now, using third equation of the motion $$v^2=v_o^2+2gx$$ $$\color{blue}{v=\sqrt{v_0^2+2gx}}$$ Again notice, $$v=\sqrt{v_o^2+2gx}$$$$=\left(v_0^2+2gx\right)^{1/2}$$ $$=v_0\left(1+\frac{2gx}{v_0^2}\right)^{1/2}$$ Now, assume that initial velocity $v_0$ is large enough then $\left|\frac{2gx}{v_0^2}\right|<1$ thus using binomial expansion of $\left(1+\frac{2gx}{v_0^2}\right)^{1/2}$ as follows $$v=v_0\left(1+\frac{\frac{1}{2}}{1!}\left(\frac{2gx}{v_0^2}\right)+\frac{\frac{1}{2}\left(\frac{1}{2}-1 \right)}{2!}\left(\frac{2gx}{v_0^2}\right)^2+\ldots\right)$$ Neglecting higher power terms, we get approximate value of velocity gain $(v-v_0)$ as follows $$v\approx v_0\left(1+\frac{1}{2}\frac{2gx}{v_0^2}\right)$$ $$v\approx v_0+\frac{gx}{v_0}$$ $$\color{blue}{v-v_0\approx\frac{gx}{v_0} }$$