A canon shoots a ball upwards with an initial speed of 100 m/s. The total time of flight of the ball is 20 s before it hits the ground. The ball loses 70% of its speed after hitting the ground. Which among the following is the correct height that the ball will bounce up after its first bounce? (g = 10 m/s²)

The ball is shot upwards with an initial speed of 100 m/s. Under gravity (g = 10 m/s²), the time to reach maximum height is 10 seconds (v/g), and the total time of flight to return to the ground is 20 seconds [1]. Upon impact, the ball's speed is 100 m/s. The problem states the ball loses 70% of its speed, meaning it retains 30% of its impact speed. The rebound velocity (v') is 30% of 100 m/s, which equals 30 m/s. To find the height (h) of the first bounce, we use the kinematic formula v² = u² + 2as. At the peak of the bounce, final velocity is 0, so 0 = (30)² - 2(10)(h). This simplifies to 20h = 900, resulting in h = 45 meters. Thus, the ball will bounce up to a height of 45 m.

Sources

1. [1] Science ,Class VIII . NCERT(Revised ed 2025) > Chapter 5: Exploring Forces > Activity 5.8: Let us observe > p. 72