Answer:
For the ball to take 1.0 s to reach the floor the shelf's height above the floor would have to be 4 h.
Explanation:
According to second equation of motion, the height or the distance covered by an object is given by :
\(s=ut+\dfrac{1}{2}at^2\)
Since, the ball rolls so the initial vertical component of velocity will be equal to 0. From above equation it is clear that,
\(h\propto t^2\)
Let h be the height above the floor when the ball taken 0.5 s to hit. At 1 s the ball is H height above the floor.
So, \(\dfrac{h}{H}=\dfrac{kt_1^2}{kt_2^2}\)
\(\dfrac{h}{H}=\dfrac{0.5^2}{1^2}\)
After solving, H = 4 h
Hence, for the ball to take 1.0 s to reach the floor the shelf's height above the floor would have to be 4 h.
