Astone is thrown vertically upward with a speed of 18.0 .
(a)how fast is it moving when it reaches a height of 11.0 ?
(b)how long is required to reach this height?

A) We can apply the equation:
2as = v² - u²
Where a is the acceleration due to gravity and opposite in direction to velocity.
v = √(2x -9.81 x 11 + 18²)
v = 10.4 m/s

b) We can apply the equation:
s = ut + 1/2 x at²
11 = 17t - 1/2 x 9.81t²
4.91t² - 17t + 11 = 0
Solving the quadratic for t,
t = 0.86; t = 2.61
The time taken to reach 11 m is 0.86 s. The larger time is when the object is falling back down and crosses the height of 11 m.

Amethod of procedure that has characterized natural science since the 17th century, consisting in systematic observation, measurement, and experiment, and the formulation, testing, and modification of hypotheses.