Students are performing an experiment for their physics class and are testing their predictions for projectile motion. They set up a system so that a beanbag is launched horizontally off one of their desks with an initial speed of 4.2 m/s. They measure the height of the desk to be 1.3 m. a) What time of flight should the students predict for the beanbag? b) What range should the students predict for the beanbag?

1. The vertical motion can be analyzed using the formula for the position under constant acceleration:
y = y_0 + v_{y0} t - \frac{1}{2} g t^2
2. Since the beanbag is launched horizontally, the initial vertical velocity ( v_{y0} ) is 0, and the equation simplifies to:
y = y_0 - \frac{1}{2} g t^2
3. Setting y = 0 when the beanbag hits the ground and using y_0 = 1.3 m:
0 = 1.3 - \frac{1}{2} (9.81) t^2
4. Solving for t :
1.3 = \frac{1}{2} (9.81) t^2
2 \times 1.3 = 9.81 t^2
2.6 = 9.81 t^2
t^2 = \frac{2.6}{9.81}
t^2 = 0.2651
t = \sqrt{0.2651}
t \approx 0.515 \; \text{seconds}

Answer: t = 0.515 \; \text{seconds}

**b) What range should the students predict for the beanbag?**

[Solution]
R = 2.163 \; \text{meters}

[Step-by-Step]
1. The horizontal motion can be analyzed using the formula for uniform motion:
x = v_x t
2. We already found the time t = 0.515 seconds.
3. Using the initial horizontal speed v_x = 4.2 m/s:
x = 4.2 \times 0.515
4. Calculating x :
x = 4.2 \times 0.515
x \approx 2.163 \; \text{meters}

Answer: R = 2.163 \; \text{meters}