a. [2 marks]

For \( H(t) = a \cos(7.8t) + b \), period of \( \cos(bt) \) is \( T = \frac{2\pi}{b} \) (M1)
\( b = 7.8 \), so \( T = \frac{2\pi}{7.8} \approx \frac{6.2832}{7.8} \approx 0.806 \) seconds (A1)
Answer: Period = \( 0.806 \) seconds (A1)

b. [3 marks]

Maximum height = 1.8 m, minimum height = 1.0 m (M1)
Amplitude: \( |a| = \frac{1.8 – 1.0}{2} = 0.4 \). Since graph starts at minimum with \( \cos \), \( a = -0.4 \) (A1)
Midline: \( b = \frac{1.8 + 1.0}{2} = 1.4 \) (A1)
Answer: \( a = -0.4 \), \( b = 1.4 \)

c. [2 marks]

Maximum when \( \cos(7.8t) = 1 \) (M1)
Period = 0.806 s. In 5 seconds: \( \frac{5}{0.806} \approx 6.2 \), so 6 complete cycles (A1)
Answer: 6 times

d. [2 marks]

Solve \( H(t) = 1.5 \): \( -0.4 \cos(7.8t) + 1.4 = 1.5 \) (M1)
\( -0.4 \cos(7.8t) = 0.1 \implies \cos(7.8t) = -0.25 \)
\( 7.8t = \cos^{-1}(-0.25) \approx 1.823 \implies t \approx \frac{1.823}{7.8} \approx 0.234 \) seconds (A1)
Answer: \( t = 0.234 \) seconds

e. [4 marks]

Find \( P(H(t) > 1.5) \): \( -0.4 \cos(7.8t) + 1.4 > 1.5 \) (M1)
\( \cos(7.8t) < -0.25 \). Solve: \( \cos^{-1}(-0.25) \approx 1.823 \) (A1)
In one period, \( \cos(7.8t) < -0.25 \) for \( \theta \in [1.823, \pi] \cup [3\pi – 1.823, 4\pi – 1.823] \)
Proportion: \( \frac{2(\pi – 1.823)}{2\pi} \approx \frac{2 \cdot 1.318}{6.283} \approx 0.406 \) (A1)
Over 5 seconds, probability remains \( 0.406 \) (A1)
Answer: Probability = \( 0.406 \)