Question

b. the thump
you will measure the speed of a thump. our thump will be made with your index finger, but many methods of thumping something are possible. you may use any method to deliver the thump: a pencil, a spring, a rubber band stretched between two fingers, but the thump must be made horizontally. we could use the fact that speed = distance/time, but because of the speeds and distances available at home, we cannot easily measure the time of movement, which is typically < 1 s. our reaction time (see previous experiments) would be a significant fraction of this time, so our results will not be reliable because we will not be able to measure the time reliably. we will instead use the drop time of our projectile from a distance h to constrain the distance traveled, x. use your knowledge of kinematic equations to determine the drop time of a projectile launched horizontally. take a projectile launched from a height h, what is the fall time of the projectile in terms of h and g? show your work below.
take a projectile launched horizontally at speed v. if the projectile travels a distance x in time t, its speed is given by v = x/t (note there is no acceleration in the horizontal direction). plugging in the time found above, determine the speed of the projectile (launched horizontally from a height h) in terms of x, h, and g. show your work below.

Explanation:

Step1: Determine the fall - time

The vertical motion of the projectile is a free - fall motion. The kinematic equation for vertical displacement $y = v_{0y}t+\frac{1}{2}at^{2}$. Since the projectile is launched horizontally, $v_{0y} = 0$, and the vertical displacement $y=-H$ (taking downwards as negative) and $a=-g$. So, $-H = 0\times t-\frac{1}{2}gt^{2}$. Solving for $t$:
\[

$$\begin{align*} -H&=-\frac{1}{2}gt^{2}\\ t^{2}&=\frac{2H}{g}\\ t&=\sqrt{\frac{2H}{g}} \end{align*}$$

\]

Step2: Determine the speed of the projectile

The horizontal speed of the projectile is given by $v=\frac{X}{t}$. Substituting $t = \sqrt{\frac{2H}{g}}$ into the formula for $v$:
\[
v = X\sqrt{\frac{g}{2H}}
\]

Answer:

The fall - time of the projectile is $t=\sqrt{\frac{2H}{g}}$, and the speed of the projectile is $v = X\sqrt{\frac{g}{2H}}$