(a) (i) METHOD 1
Attempt to find change in height of the ball using gradient (M1)
\( \frac{a}{0.43} = -0.045 \)
\( a = -0.045 \times 0.43 \)
\( a = -0.01935 \) (m) (\(-0.0194\) (m)) (A1)
Result: \( a = -0.0194 \) m [2]

METHOD 2
Attempt to find height at back of home plate (M1)
Horizontal distance to the front of home plate: \( y = 1.25 = -0.045x + 2 \implies -0.045x = -0.75 \implies x = \frac{0.75}{0.045} \approx 16.6667 \) m
Distance to back of home plate: \( 16.6667 + 0.43 = 17.0967 \) m
Height at back: \( y = -0.045 \times 17.0967 + 2 \approx -0.76935 + 2 = 1.23065 \) m
\( a = 1.23065 – 1.25 = -0.01935 \) (m) (\(-0.0194\) (m)) (A1)
Result: \( a = -0.0194 \) m [2]

(ii)
Height at back of home plate: \( 1.23065 \) m (from (i)) (M1)
Strike zone: \( 0.53 < 1.23065 < 1.24 \) (A1)
Therefore, the throw is a strike (R1)
Result: The throw is a strike because the height at the back of home plate (1.23065 m) is between 0.53 m and 1.24 m [2]

(b) METHOD 1
Indication of \( d = 96 \) in the function \( h(d) \) or its graph (M1)
\( h(96) = -0.01(96)^2 + 1.04 \cdot 96 + 0.66 \) (A1)
\( h(96) = -0.01 \cdot 9216 + 99.84 + 0.66 = -92.16 + 99.84 + 0.66 = 8.34 \) (m) (A1)
\( 8.34 > 5 \), so the ball will go over the wall (A1)
Result: The ball goes over the wall because its height at 96 m (8.34 m) exceeds the wall height (5 m) [2]

METHOD 2
Indication of \( h = 5 \) in the function \( h(d) \) or its graph (M1)
Solve: \( 5 = -0.01d^2 + 1.04d + 0.66 \) (A1)
\( -0.01d^2 + 1.04d + 0.66 – 5 = 0 \implies -0.01d^2 + 1.04d – 4.34 = 0 \)
\( d^2 – 104d + 434 = 0 \)
Discriminant: \( \Delta = (-104)^2 – 4 \cdot 1 \cdot 434 = 10816 – 1736 = 9080 \)
\( d = \frac{104 \pm \sqrt{9080}}{2} \approx \frac{104 \pm 95.26}{2} \)
\( d \approx 99.63 \) or \( d \approx 4.37 \) (m) (A1)
Since \( 96 < 99.63 \), the ball reaches 5 m after 96 m, so it is above 5 m at \( d = 96 \), thus it goes over the wall (A1)
Result: The ball goes over the wall because it is above 5 m at \( d = 96 \) m [2]