#### Answer

(a) The time to reach the lowest point is 1.28 seconds.
(b) The time to reach the lowest point for the second time is 3.84 seconds.

#### Work Step by Step

(a) We can find the period of the oscillation.
$T = 2\pi~\sqrt{\frac{L}{g}}$
$T = 2\pi~\sqrt{\frac{6.50~m}{9.80~m/s^2}}$
$T = 5.117~s$
The climber reaches the lowest point after completing one-fourth of a cycle. We can find the time to reach the lowest point.
$t = \frac{T}{4} = \frac{5.117~s}{4} = 1.28~s$
The time to reach the lowest point is 1.28 seconds.
(b) The climber reaches the lowest point for the second time after completing three-fourths of a cycle. We can find the time to reach the lowest point for the second time.
$t = \frac{3T}{4} = \frac{(3)(5.117~s)}{4} = 3.84~s$
The time to reach the lowest point for the second time is 3.84 seconds.