EXERCISE 11.1 Solve the following problems: 1. A flowerpot falls from a window sill 30m above the street. a. How long does it take for the flowerpot to hit the street? b. What is its velocity when it hits the street? 2. A tennis ball is tossed upward. If it moves 1.5 m before it starts to fall down again, with what speed was it thrown? 3. A juggler throws a bowling pin upward with an initial velocity of 9.8 m/s. a. How long does it take for the bowling pin to reach its highest point ?b. When it returns to the juggler's hand (assuming it's stil where it was when the bowling pin was released), what i its velocity? 4. A tool bag falls from a height of 10 m at a construction site.a. What is its velocity after 1 s? b. How far has it traveled after 1 s?c. How long does it take to reach the ground?d. With what velocity does it hit the ground?
Answer (1)
Answer:Here are the solutions to the following projectile motion problems. We'll use g = 9.8 m/s² for the acceleration due to gravity. Problem 1: - a. Time to hit the street: We use the equation: d = v₀t + (1/2)at² Where: - d = 30 m (distance) - v₀ = 0 m/s (initial velocity) - a = 9.8 m/s² (acceleration due to gravity) - t = time (what we're solving for) 30 m = 0 + (1/2)(9.8 m/s²)t²t² = (30 m) / (4.9 m/s²)t² ≈ 6.12 s²t ≈ √6.12 s² ≈ 2.47 s - b. Velocity when it hits the street: We use the equation: v = v₀ + at v = 0 + (9.8 m/s²)(2.47 s)v ≈ 24.2 m/s Problem 2: - Speed when thrown: At the highest point, the velocity is 0. We use: v² = v₀² + 2ad Where: - v = 0 m/s (final velocity at highest point) - v₀ = initial velocity (what we're solving for) - a = -9.8 m/s² (negative because gravity acts downwards) - d = 1.5 m (distance) 0 = v₀² + 2(-9.8 m/s²)(1.5 m)v₀² = 29.4 m²/s²v₀ = √29.4 m²/s² ≈ 5.42 m/s Problem 3: - a. Time to reach highest point: At the highest point, the final velocity is 0. We use: v = v₀ + at Where: - v = 0 m/s (final velocity) - v₀ = 9.8 m/s (initial velocity) - a = -9.8 m/s² (negative because gravity acts downwards) - t = time (what we're solving for) 0 = 9.8 m/s + (-9.8 m/s²)tt = (9.8 m/s) / (9.8 m/s²) = 1 s - b. Velocity when it returns: The time to go up equals the time to come down (assuming no air resistance). The velocity will be equal in magnitude but opposite in direction to the initial velocity. Therefore, the velocity when it returns is -9.8 m/s. Problem 4: - a. Velocity after 1 s: v = v₀ + atv = 0 + (9.8 m/s²)(1 s) = 9.8 m/s - b. Distance traveled after 1 s: d = v₀t + (1/2)at²d = 0 + (1/2)(9.8 m/s²)(1 s)² = 4.9 m - c. Time to reach the ground: 10 m = (1/2)(9.8 m/s²)t²t² = (10 m) / (4.9 m/s²)t² ≈ 2.04 s²t ≈ √2.04 s² ≈ 1.43 s - d. Velocity when it hits the ground: v = v₀ + atv = 0 + (9.8 m/s²)(1.43 s) ≈ 14.0 m/s Remember that these calculations ignore air resistance. In reality, air resistance would affect the results, especially for the flowerpot and the tool bag.
