A person is stranded and sitting 45m above the ground on the ledge of the cliff. of a safety rope is launched upward at an angle of 50 degrees above the horizontal at a distance of 60m away, what is the initial velocity of the rope so that it reaches the stranded person?

Answer 1

Y^2 = Yo^2 + 2g*h = 0 @ max ht.
Yo^2 = - 2(-9.8)*45 = 882
Yo = 29.7 m/s = Vertical component of
initial velocity.

Vo*sin50 = Yo = 29.7
Vo = 38.8 m/s.