Answer:
d = 1.02 m
Explanation:
By energy conservation we can find the speed by which ball will leave the ramp
[tex]U_i + KE_i = U_f + KE_f[/tex]
here we know that
[tex]mgh_1 + 0 = mgh_2 + \frac{1}{2}mv^2[/tex]
here we have
[tex]h_1 = 1.17 m[/tex]
[tex]h_2 = 0.298 m[/tex]
so we have
[tex](9.8)(1.17) = (9.8)(0.298) + \frac{1}{2} v^2[/tex]
[tex]v = 4.13 m/s[/tex]
now the time taken by the block to reach the ground is given by
[tex]h_2 = \frac{1}{2}gt^2[/tex]
[tex]t = \sqrt{\frac{2h_2}{g}}[/tex]
[tex]t = \sqrt{\frac{2(0.298)}{9.8}}[/tex]
[tex]t = 0.25 s[/tex]
now the distance covered by it is given as
[tex]d = 0.25 \times 4.13[/tex]
[tex]d = 1.02 m[/tex]
