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 looses 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)
100 m
70 m
50 m
45 m
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC CAPF – 2024
The ball is shot upwards with an initial speed of 100 m/s. In vertical motion under gravity (g=10 m/s²), the time taken to reach the maximum height is given by t_up = u/g = 100/10 = 10 s. The time taken to fall back to the ground from the maximum height is also 10 s. The total time of flight before the first bounce is t_up + t_down = 10 + 10 = 20 s, which matches the given information. The speed of the ball just before hitting the ground after falling from its peak height is equal to its initial projection speed, which is 100 m/s (downwards). After hitting the ground, the ball loses 70% of its speed. So, the speed after the first bounce is (100% – 70%) of 100 m/s = 30% of 100 m/s = 0.30 * 100 m/s = 30 m/s (upwards). To find the height the ball bounces up, we use the equation v² = u² + 2as, where the final velocity at the peak height is v=0, the initial velocity after bounce is u=30 m/s, and the acceleration is a=-g=-10 m/s². So, 0² = 30² + 2 * (-10) * h. This simplifies to 0 = 900 – 20h, so 20h = 900, which gives h = 900/20 = 45 meters.
Understanding projectile motion under gravity, calculating time of flight, speed before impact, calculating speed after an inelastic collision (loss of speed), and calculating the maximum height reached with a new initial velocity.