When the ball is dropped, there is a gravitational force of attraction acting on it that accelerates it towards the ground. The average acceleration due to gravity on the Earth is 9.8 m/s^2.
If a body starts with velocity u m/s and is accelerated at a m/s, the distance traveled by it in time t seconds is equal to D = u*t + (1/2)*a*t^2
In the problem, the ball is dropped from a height of 18 m. Its initial velocity is 0 m/s. The time taken by the ball to fall 18 m is t where 18 = 0*t + (1/2)*9.8*t^2
t = `sqrt(18/4.9)` = 1.916 s
The ball takes 1.916 s to strike the ground.
After striking the ground, the ball bounces up and rises to a height 40% of the height it is dropped from. After the first bounce it reaches 7.2 m. When it falls again, the total distance traveled when it strikes the ground is 18+7.2*7.2 = 32.4 m. The ball has traveled 30 m before it reaches the ground the second time.
It travels the first 18 m in 1.916 s. As the ball only rises to 40% of its height, the velocity with which it starts in the upward direction after striking the ground the first time is u = `sqrt(2*9.8*7.2)` = 11.879 m/s. The time taken by the ball to rise up to 7.2 m is equal to 1.21 s. The time taken by the ball to travel the last 4.8 m 0.9897 s. The total time taken by the ball to travel 30 m is 4.1157 s.
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