Question

scenario
an engineer is testing the design of a new crossbow to try to find the speed of a dart the instant it is launched from the crossbow. the crossbow is fixed to a table so that the dart is launched horizontally. a target is placed in front of the crossbow so that the crossbow is pointing directly at the center of the target. the distance d from the crossbow to the target is varied, and each time the crossbow is fired, the distance h from the center that the dart strikes the target is measured.
quantitative analysis
the data taken by the engineer are shown in the table.
horizontal distance to target d(m) vertical distance dart falls h(m) time the dart is a projectile t(s)
5.00 0.084 0.131
10.00 0.169 0.185*
15.00 0.408 0.2866
20.00 0.677 0.371
25.00 0.975 0.446
part a: using the assumption that the crossbow darts were projected horizontally, derive an expression for t in terms of h and g. (remember that derivations should start with a fundamental equation of physics, i.e., an equation given on the ap physics 1 equation sheet.)
step 1: h = 1/2gt^2
step 2: 2h/g = t
step 3:
step 4:
part b: fill in the data table with five values of t calculated using your expression in part a. use g = 9.8 m/s^2 and a reasonable number of significant figures.

Explanation:

Step1: Recall vertical - motion equation

The vertical - displacement of an object in free - fall is given by $H = \frac{1}{2}gT^{2}$, where $H$ is the vertical distance, $g$ is the acceleration due to gravity, and $T$ is the time of flight.

Step2: Solve for $T$

Starting with $H=\frac{1}{2}gT^{2}$, we first multiply both sides of the equation by $2$ to get $2H = gT^{2}$. Then, we divide both sides by $g$: $T^{2}=\frac{2H}{g}$. Taking the square - root of both sides, we have $T=\sqrt{\frac{2H}{g}}$.

For Part B:

1. When $H = 0.084m$ and $g = 9.8m/s^{2}$, $T=\sqrt{\frac{2\times0.084}{9.8}}\approx0.131s$.
2. When $H = 0.169m$, $T=\sqrt{\frac{2\times0.169}{9.8}}\approx0.186s$.
3. When $H = 0.408m$, $T=\sqrt{\frac{2\times0.408}{9.8}}\approx0.288s$.
4. When $H = 0.677m$, $T=\sqrt{\frac{2\times0.677}{9.8}}\approx0.370s$.
5. When $H = 0.975m$, $T=\sqrt{\frac{2\times0.975}{9.8}}\approx0.446s$.

Answer:

For Part A: $T=\sqrt{\frac{2H}{g}}$
For Part B:

Horizontal Distance to Target $D(m)$	Vertical Distance Dart Falls $H(m)$	Time the Dart Is a Projectile $T(s)$
10.00	0.169	0.186
15.00	0.408	0.288
20.00	0.677	0.370
25.00	0.975	0.446