(a) METHOD 1
Attempt to find vertex of quadratic \( h(t) = -4.75t^2 + 8.75t + 1.5 \) using \( t = -\frac{b}{2a} \) (M1)
Coefficients: \( a = -4.75 \), \( b = 8.75 \)
\( t = -\frac{8.75}{2 \cdot (-4.75)} = \frac{8.75}{9.5} \approx 0.921052 \) seconds (A1)
Result: 0.921 seconds [2]

METHOD 2
Use derivative to find maximum: \( h'(t) = -9.5t + 8.75 = 0 \) (M1)
Solve: \( -9.5t = -8.75 \implies t = \frac{8.75}{9.5} \approx 0.921052 \) seconds (A1)
Result: 0.921 seconds [2]

(b) METHOD 1
Set \( h(t) = 0 \): \( -4.75t^2 + 8.75t + 1.5 = 0 \) (M1)
Quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), where \( a = -4.75 \), \( b = 8.75 \), \( c = 1.5 \)
Discriminant: \( 8.75^2 – 4 \cdot (-4.75) \cdot 1.5 = 76.5625 + 28.5 = 105.0625 \)
\( t = \frac{-8.75 \pm \sqrt{105.0625}}{-9.5} \approx \frac{-8.75 \pm 10.25}{-9.5} \)
Roots: \( t \approx -0.1579 \) (discard, \( t \geq 0 \)), \( t = 2 \) seconds (A1)
Result: 2 seconds [2]

METHOD 2
Factorize: \( -4.75t^2 + 8.75t + 1.5 = 0 \implies t^2 – \frac{8.75}{4.75}t – \frac{1.5}{4.75} = 0 \implies t^2 – \frac{35}{19}t – \frac{6}{19} = 0 \) (M1)
Solve: \( t = \frac{\frac{35}{19} \pm \sqrt{\left(\frac{35}{19}\right)^2 – 4 \cdot 1 \cdot \left(-\frac{6}{19}\right)}}{2} \)
Discriminant: \( \frac{1225}{361} + \frac{24}{19} = \frac{200}{19} \), so \( t \approx \frac{\frac{35}{19} \pm \frac{\sqrt{200}}{19}}{2} \)
Roots: \( t \approx -0.1579 \), \( t = 2 \) seconds (A1)
Result: 2 seconds [2]

(c) METHOD 1
Set \( h(t) = 1.2 \): \( -4.75t^2 + 8.75t + 1.5 = 1.2 \implies -4.75t^2 + 8.75t + 0.3 = 0 \) (M1)
Quadratic formula: \( a = -4.75 \), \( b = 8.75 \), \( c = 0.3 \)
Discriminant: \( 8.75^2 – 4 \cdot (-4.75) \cdot 0.3 = 76.5625 + 5.7 = 82.2625 \)
\( t = \frac{-8.75 \pm \sqrt{82.2625}}{-9.5} \approx \frac{-8.75 \pm 9.07}{-9.5} \)
Roots: \( t \approx -0.0337 \) (discard), \( t \approx 1.8758 \approx 1.88 \) seconds (A1)
Result: 1.88 seconds [2]

METHOD 2
Factorize: \( -4.75t^2 + 8.75t + 0.3 = 0 \implies t^2 – \frac{8.75}{4.75}t – \frac{0.3}{4.75} = 0 \implies t^2 – \frac{35}{19}t – \frac{6}{95} = 0 \) (M1)
Solve: \( t = \frac{\frac{35}{19} \pm \sqrt{\left(\frac{35}{19}\right)^2 – 4 \cdot 1 \cdot \left(-\frac{6}{95}\right)}}{2} \)
Discriminant: \( \frac{1225}{361} + \frac{24}{95} \approx 3.645 \), so \( t \approx \frac{\frac{35}{19} \pm \sqrt{3.645}}{2} \)
Roots: \( t \approx -0.0337 \), \( t \approx 1.8758 \approx 1.88 \) seconds (A1)
Result: 1.88 seconds [2]

(d)
The model ignores air resistance (R1)
Note: Accept other valid limitations, e.g., assumes ball as a point, constant gravity, no bouncing after hitting ground, or ignores spin/wind effects.
Result: The model ignores air resistance, which would affect the trajectory of the basketball [1]