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.
