A projectile launched horizontally passed through two photogates separated by 38.6 cm. The measured time for the projectile to pass through the two photogates was 59 ms. Calculate the velocity of the projectile in m/s.

The question required the calculation of the velocity of a horizontally launched projectile that passed through two photogates. By using the distance between the photogates and the time it took to pass through them, the velocity was calculated using the formula v = s/t. The calculated velocity is approximately 6.54 m/s.

Explanation:

The subject of this question is physics, and it relates specifically to projectile motion and velocity calculation. In this scenario, a projectile is launched horizontally and passes through two photogates separated by 38.6 cm. The time it took for this projectile to pass through the photogates was recorded at 59 ms or 0.059 seconds.

The task is to calculate the velocity of the projectile. Velocity is calculated by dividing the displacement (distance traveled) by the time it takes to travel that distance. Here, the distance traveled (displacement) by the projectile is 38.6 cm or 0.386 meters (since we need the distance in the standard SI unit of meter).

So, using the formula of velocity (v) = displacement (s)/time (t), we substitute the given values as v = s/t = 0.386m / 0.059s which approximately equals 6.54m/s. Therefore, the velocity of the projectile can be calculated as roughly 6.54 meters per second.

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