A circus performer juggling while standing on a tightrope suspended 18.0 m above the ground tosses a sword directly upward into the air with a speed of 5.0 m/s. If the sword leaves his hand 1.0 m above the tightrope, what is the sword's maximum height above the ground?

Explanation:

let's use kinematics to solve for max height above the hand.

At the top of its arc, the sword has zero vertical velocity.

v² = u² + 2as

s = (v² - u²)/2a

s = (0² - 5.0²) / (2(-9.8))

s = 1.2755...m

call it 1.3 m

to this we must adjust for the height of the hand at release which is

18.0 + 1.0 = 19.0 m above ground

19.0 + 1.3 = 20.3 m above ground