Horizontally Launched Projectiles

Name Date Horizontally Launched Projectiles 1. Given the following situation of a marble in motion on a rail with negligible friction: a. Sketch a motion map showing the motion of the marble after it leaves the rail, using the grid to help you carefully locate the marble’s positions. Show both horizontal and vertical velocity vectors on each dot. v = 10.0 m/s h = 1.5 m b. Once the ball leaves the table, calculate how long it will take for the ball to hit the floor. • Using: Δy = v₀t + ½gt² • -1.5 = 0 + ½(-9.8)t² • t² = 1.5/4.9 = 0.306 • t ≈ 0.55 seconds Answer: t ≈ 0.55 seconds c. In the time you have calculated in part b, how far will the ball travel horizontally before hitting the floor? • Δx = vₓt = 10.0 m/s × 0.55 s • Δx ≈ 5.5 meters Answer: Δx ≈ 5.5 meters d. Suppose the table was doubled in height to 3.0 m. Determine the horizontal range of the marble as it falls to the floor. What effect does doubling the height have on range of the marble? What other factors affect the range of the marble? 3.0 = ½(9.8)t² t² = 0.612 t ≈ 0.78 s Δx = 10.0 × 0.78 ≈ 7.8 meters Range = 7.8 meters; doubling height increases range by factor of √2 1 2. A student finds that it takes 0.20s for a ball to pass through photogates placed 30 cm apart on a level ramp. The end of the ramp is 92 cm above the floor. Where could a coin be placed so that the ball directly strikes the coin on impact with the ground? photogates table • vₓ = distance/time = 0.30 m / 0.20 s = 1.5 m/s Time to fall from 0.92 m: • 0.92 = ½(9.8)t² • t² = 0.188 • t ≈ 0.433 s floor Horizontal distance from end of ramp: • Δx = vₓ × t = 1.5 × 0.433 ≈ 0.65 m The coin should be placed 65 cm (0.65 m) from the end of the ramp 3. Suppose now that the same ball, released from the same ramp (92 cm high) struck a coin on the floor placed 25 cm from the end of the ramp. a. What was the ball's horizontal velocity? Time to fall 0.92 m: • 0.92 = ½(9.8)t² • t ≈ 0.433 s Since it travels 0.25 m horizontally: • vₓ = distance/time = 0.25 / 0.433 • vₓ ≈ 0.58 m/s Answer: vₓ ≈ 0.58 m/s b. How long did it take for the ball to pass through the photogates? If the photogates are still 30 cm apart: • Δt = distance/velocity = 0.30 m / 0.58 m/s • Δt ≈ 0.52 seconds Answer: Δt ≈ 0.52 seconds 2