As suggested by @Kulambo's answer and @Farcher's comment a pendulum is a great method.

The standard period of a pendulum $T = 2\pi \sqrt{L/g}$ depends on the small angle approximation, but is independent of the mass of the pendulum bob. The uncertainty in your determination of $g$ will depend on how well you can measure $L$ and $T$.

$$\delta g = g \sqrt{ \left(\frac{\delta L}{L}\right)^2 + 2\left(\frac{\delta T}{T}\right)^2}$$.

The small angle approximation $\sin\theta \approx \theta$ is good to about $2$% for $\theta = 20^\circ = 0.35$ rad. So keep your swing angle smaller than that.

Remember $L$ is the length from the pivot to the center of mass of the pendulum. With a light string and a heavy, symmetric bob this will be very close to the geometric center of the bob. It is better to fix the string in place with a clamp, instead of tying it. This way it won't slide, so there won't be much frictional loss at the pivot. It's also easier to precisely locate the true pivot.

To better determine $T$, you can take the average of many swings. The biggest uncertainty will be your ability to press start and stop on your stopwatch at the correct times. If you time the pendulum making five complete swings then divide by five, that effectively cuts the start/stop error by a factor of five. Frictional losses will noticeably alter the period if you let it go for too long. So there is a trade off in how many swings to average over.

A $1$ m pendulum should have a period of about $2$ sec. Lets say you determine the pivot to center of mass length to $1$ cm. Then you time five swings ($10$ sec) with $0.5$ sec of start/stop uncert. This will get you a determination of $g$ with better than $10$% precision.

$$\delta g = (9.8\,\mathrm{m/s}^2) \, \sqrt{\left( \frac{0.01\,\mathrm{m}}{1.00\,\mathrm{m}} \right)^2 + 2 \left( \frac{0.1\,\mathrm{s}}{2.0\,\mathrm{s}} \right)^2} \approx 0.7 \,\mathrm{m/s}^2$$

The limiting factor will likely be your ability to use the stop watch. You could do much better using your photogate. If you can get down to $0.01$ sec of start/stop precision on one swing, the measurement uncertainty will get close to $1$%. This would make the systematic error from the small angle approximation in the pendulum model take over as the dominate source of uncertainty. The lack of friction and air resistance in the model may also be a factor at the $1$% level.

As suggested by @Kulambo's answer and @Farcher's comment a pendulum is a great method. There's a long history of pendulum gravimeters dating back to the 1600's.

The standard period of a pendulum $T = 2\pi \sqrt{L/g}$ depends on the small angle approximation, but is independent of the mass of the pendulum bob. The uncertainty in your determination of $g$ will depend on how well you can measure $L$ and $T$.

$$\delta g = g \sqrt{ \left(\frac{\delta L}{L}\right)^2 + 2\left(\frac{\delta T}{T}\right)^2}$$.

The small angle approximation $\sin\theta \approx \theta$ is good to about $2$% for $\theta = 20^\circ = 0.35$ rad. So keep your swing angle smaller than that.

Remember $L$ is the length from the pivot to the center of oscillation of the pendulum. With a light string and a small, heavy bob this will be very close to the center of mass of the bob. It is better to fix the string in place with a clamp, instead of tying it. This way it won't slide, so there won't be much frictional loss at the pivot. It's also easier to precisely locate the true pivot.

To better determine $T$, you can take the average of many swings. The biggest uncertainty will be your ability to press start and stop on your stopwatch at the correct times. If you time the pendulum making five complete swings then divide by five, that effectively cuts the start/stop error by a factor of five.

A $1$ m pendulum should have a period of about $2$ sec. Lets say you determine the pivot to center of oscillation length, $L$, to $1$ cm. Then you time five swings ($5T = 10$ sec) with $0.5$ sec of start/stop uncert. This will get you a determination of $g$ with better than $10$% precision.

$$\delta g = (9.8\,\mathrm{m/s}^2) \, \sqrt{\left( \frac{0.01\,\mathrm{m}}{1.00\,\mathrm{m}} \right)^2 + 2 \left( \frac{0.1\,\mathrm{s}}{2.0\,\mathrm{s}} \right)^2} \approx 0.7 \,\mathrm{m/s}^2$$

The limiting factor will likely be your ability to use the stop watch. You could do much better using your photogate. If you can get down to $0.01$ sec of start/stop precision on one swing, the measurement uncertainty will get close to $1$%. This would make the systematic error from the small angle approximation in the pendulum model take over as the dominate source of uncertainty.

The lack of friction and air resistance in the model may also be a factor if you are trying to get to the $1$% level. Historically, precision pendulums were operated in evacuated, low pressure chambers to reduce air resistance. Candela et al 2000 claim that one can measure $g$ to parts in $10^4$ using a Bessel pendulum. They also discuss how combining data from both pivots of a reversible pendulum causes the air resistance errors to cancel when calculating $g$. (The article is paywalled, but you can probably get it with the help of your local library.)

As suggested by @Kulambo's answer and @Farcher's comment a pendulum is a great method.

The standard period of a pendulum $T = 2\pi \sqrt{L/g}$ depends on the small angle approximation, but is independent of the mass of the pendulum bob. The uncertainty in your determination of $g$ will depend on how well you can measure $L$ and $T$.

$$\delta g = g \sqrt{ \left(\frac{\delta L}{L}\right)^2 + 2\left(\frac{\delta T}{T}\right)^2}$$.

The small angle approximation $\sin\theta \approx \theta$ is good to about $2$% for $\theta = 20^\circ = 0.35$ rad. So keep your swing angle smaller than that.

Remember $L$ is the length from the pivot to the center of mass of the pendulum. With a light string and a heavy, symmetric bob this will be very close to the geometric center of the bob. It is better to fix the string in place with a clamp, instead of tying it. This way it won't slide, so there won't be much frictional loss at the pivot. It's also easier to precisely locate the true pivot.

To better determine $T$, you can take the average of many swings. The biggest uncertainty will be your ability to press start and stop on your stopwatch at the correct times. If you time the pendulum making five complete swings then divide by five, that effectively cuts the start/stop error by a factor of five. Frictional losses will noticeably alter the period if you let it go for too long. So there is a trade off in how many swings to average over.

A $1$ m pendulum should have a period of about $2$ sec. Lets say you determine the pivot to center of mass length to $1$ cm. Then you time five swings ($10$ sec) with $0.5$ sec of start/stop uncert. This will get you a determination of $g$ with better than $10$% precision.

$$\delta g = (9.8\,\mathrm{m/s}^2) \, \sqrt{\left( \frac{0.01\,\mathrm{m}}{1.00\,\mathrm{m}} \right)^2 + 2 \left( \frac{0.1\,\mathrm{s}}{2.0\,\mathrm{s}} \right)^2} \approx 0.7 \,\mathrm{m/s}^2$$

The limiting factor will likely be your ability to use the stop watch. If you can get $0.05$ sec of start/stop precision on your five swings, the measurement uncertainty will get close to $1$%, $\approx 0.1 \,\mathrm{m/s}^2$. This would make the systematic error from the small angle approximation in the pendulum model take over as the dominate source of uncertainty.

As suggested by @Kulambo's answer and @Farcher's comment a pendulum is a great method.

The standard period of a pendulum $T = 2\pi \sqrt{L/g}$ depends on the small angle approximation, but is independent of the mass of the pendulum bob. The uncertainty in your determination of $g$ will depend on how well you can measure $L$ and $T$.

$$\delta g = g \sqrt{ \left(\frac{\delta L}{L}\right)^2 + 2\left(\frac{\delta T}{T}\right)^2}$$.

The small angle approximation $\sin\theta \approx \theta$ is good to about $2$% for $\theta = 20^\circ = 0.35$ rad. So keep your swing angle smaller than that.

Remember $L$ is the length from the pivot to the center of mass of the pendulum. With a light string and a heavy, symmetric bob this will be very close to the geometric center of the bob. It is better to fix the string in place with a clamp, instead of tying it. This way it won't slide, so there won't be much frictional loss at the pivot. It's also easier to precisely locate the true pivot.

To better determine $T$, you can take the average of many swings. The biggest uncertainty will be your ability to press start and stop on your stopwatch at the correct times. If you time the pendulum making five complete swings then divide by five, that effectively cuts the start/stop error by a factor of five. Frictional losses will noticeably alter the period if you let it go for too long. So there is a trade off in how many swings to average over.

A $1$ m pendulum should have a period of about $2$ sec. Lets say you determine the pivot to center of mass length to $1$ cm. Then you time five swings ($10$ sec) with $0.5$ sec of start/stop uncert. This will get you a determination of $g$ with better than $10$% precision.

$$\delta g = (9.8\,\mathrm{m/s}^2) \, \sqrt{\left( \frac{0.01\,\mathrm{m}}{1.00\,\mathrm{m}} \right)^2 + 2 \left( \frac{0.1\,\mathrm{s}}{2.0\,\mathrm{s}} \right)^2} \approx 0.7 \,\mathrm{m/s}^2$$

The limiting factor will likely be your ability to use the stop watch. You could do much better using your photogate. If you can get down to $0.01$ sec of start/stop precision on one swing, the measurement uncertainty will get close to $1$%. This would make the systematic error from the small angle approximation in the pendulum model take over as the dominate source of uncertainty. The lack of friction and air resistance in the model may also be a factor at the $1$% level.