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Megha
Megha
Asked: January 20, 20222022-01-20T18:28:14+05:30 2022-01-20T18:28:14+05:30In: Education and learning

Answer the following questions

On September 26,1993, Dave Munday, a diesel mechanic by trade, went over the Canadian edge of Niagara Falls for the second time, freely falling 48 m to the water (and rocks) below. On this attempt, he rode in a steel ball with a hole for air. Munday, keen on surviving this plunge that had killed four other stuntmen, had done considerable research on the Physics and engineering aspects of the plunge. During this free fall he fell 48 m. Assuming his initial velocity was zero during the free fall and neglecting the effect of the air on the ball during the fall answer the following questions. (During free fall consider acceleration due to gravity as -9.8 m/s2)

1) How long did Munday fall to reach the water surface below the falls?
2) What was Munday’s velocity as he reached the water surface?
3) What was Munday’s velocity at each count of 1 s (i.e after a period of 1 s)?
4) When do we say that an object undergoes retardation or deceleration?

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    1. Wish Lay
      2022-01-20T18:42:58+05:30Added an answer on January 20, 2022 at 6:42 pm

      Hi Megha, Let’s answer the all questions step by step

      First, note the Data Given:

      1. Depth of the fall = 48m
      2. Value of g= 9.8 m/s²
      3. Initial Velocity = 0m/s

      A) To Find How long did Munday fall to reach the water surface below the falls?

      Here we can use 2nd equation of motion since the given acceleration is constant, 

      Let s = ut + 1/2 a t²

      Since u = 0 m/s and s = 48m

      -48 = 1/2 (-9.8) t²

      or t = 3.12 s

      B) What was Munday’s velocity as he reached the water surface?

      By using 1st equation of motion we can solve the velocity

      Let v = u + at

      Substituting the value of t from (A)

      v = 0 + (-9.8) (3.12)

      you will get v = – 30.57m/s

       

      C) What was Munday’s velocity after a period of 1s?

      To find velocity after period of 1 s, we can use the 1st equation of motion.

      So let v1 = u + at

      let t = 1 s

      v1 = 0 + (-9.8) (1)

      Hence v1 = -9.8m/s

      D) When do we say that an object undergoes retardation?

      Answer: Object is said to be undergoing negative acceleration (aka retardation).

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