Answer:
v = 35.08 m/s
Explanation:
A ball rolls over the edge of a platform with a horizontal velocity of magnitude v.
The horizontal range of the ball from the base of the platform is, x = 20 m
Horizontal velocity,
[tex]v=\dfrac{x}{t}[/tex]
[tex]v=\dfrac{20}{t}[/tex]..............(1)
Vertical distance, y = 1.6 m
Vertical distance is given by :
[tex]y=ut+\dfrac{1}{2}gt^2[/tex]
u = 0
[tex]1.6=\dfrac{1}{2}\times 9.8t^2[/tex]
t = 0.57 seconds
Put the value of t in equation (1) as :
[tex]v=\dfrac{20}{0.57}[/tex]
v = 35.08 m/s
So, the horizontal velocity of the ball is 35.08 m/s. Hence, this is the required solution.
