Question

ans the question in attachment if u can​

Answer 1

Answer:

m g h = 1/2 m v^2     to get speed at point V   (conservation of energy)

v^2 = 2 g h = 100     using 10 m/s^2 for g

ac = v^2 / R       centripetal acceleration needed to remain on slope

ac = 100 / 5 = 20  > g = 10    so mass leaves curve

R = 1/2 g t^2     where t is time to fall to ground

t^2 = 2 * 5 / 10      and t = 1 sec

Horizontal speed at V:   v^2 = 2 g h = 100 from above and v = 10 m/s

So S = vx * t = 10 * 1 = 10 m from point V or 5 m from curve meets ground.