You and your friends find a rope that hangs down 15 m from a high tree branch right at the edge of a river. You find that you can run, grab the rope, swing out over the river, and drop into the water. You run at 2.0 m/s and grab the rope, launching yourself out over the river.

Vertical angle=9.5°

Time to reach maximum height = 1.24seconds

Step-by-step explanation:

The maximum distance x is reached when rope swings to a maximum height, maximum height is gotten from the equation KE= 1/2MU^2 = MgH = MgL= Mgl(1- cos theta.

Where theta is the vertical angle, U=2m/s and L= 15m

2gl -2glcostheta=U^2

2gl U^2=2gl cos theta

But theta= Acos(1- U^2)/2gl

Acos(1-/(2×9.8×15)

Theta =9.5°

Maximum height li given by

Lsintheta

But X= UxT

T=X/UX= Lsintheta/U = 15sin9.5 /2 =1.24seconds