Case Study
Passage with linked questions
Case Set 1
Case AnalysisPassage
During a sports meet, a shot-putter launches a metal ball at an initial speed of 14 m/s at an angle of 45° above the horizontal from a height close to ground level. The coach observes that the ball follows a curved path before landing. The physics teacher accompanying the team explains that the motion of the ball is a classic example of projectile motion studied in Chapter 3. She draws the trajectory on a whiteboard, marks the highest point, and asks students to calculate the maximum height reached, the total time in the air, and the horizontal distance covered. She also reminds them that air resistance is neglected in the textbook treatment, and that Galileo first described the independence of horizontal and vertical components of projectile motion in his book Dialogue on the Great World Systems in 1632.
Question 1: State the nature of acceleration acting on the shot-put ball after it leaves the athlete's hand, and write its components along x and y directions.
- After projection, the only acceleration acting on the ball is due to gravity, directed vertically downward with magnitude g.
- The horizontal component of acceleration ax = 0, since no force acts in the horizontal direction (air resistance neglected).
- The vertical component ay = −g (taking upward as positive), so the vertical velocity decreases on the way up and increases on the way down.
Question 2: Calculate the maximum height reached by the ball and the total time of flight. (g = 9.8 m/s²)
- At 45°: v0 sin θ0 = 14 × sin 45° = 14/√2 ≈ 9.9 m/s.
- Maximum height hm = (v0 sin θ0)²/(2g) = (9.9)²/(2 × 9.8) = 98.01/19.6 ≈ 5.0 m.
- Time of flight Tf = 2v0 sin θ0/g = 2 × 9.9/9.8 ≈ 2.02 s.
Question 3: Calculate the horizontal range and explain why 45° gives the maximum range for a given initial speed.
- Range R = v0² sin 2θ0/g = (14)² × sin 90°/9.8 = 196/9.8 = 20 m.
- R is maximum when sin 2θ0 = 1, i.e., 2θ0 = 90°, so θ0 = 45°; at this angle the product of horizontal speed and time of flight is maximised.
- For θ0 < 45°, the time of flight is short; for θ0 > 45°, the horizontal speed component is small — at 45° the balance between these two factors gives the greatest range Rm = v0²/g.