**7. (a)**
$1 = 12 \sin 50°t - 4.9t^2$, $4.9t^2 - 12 \sin 50°t + 1 = 0$, $t = 0.116$ or $t = 1.76$, $t = 1.76 \text{ s}$ | M1A1; dM1, A1; (M2A1A1) | M1: Seeing three terms equation with: 12 $\sin50°$ or 12 $\cos50°$ and 1 and $\pm 4.9t^2$. A1: Correct terms with possible sign errors. dM1: At least one solution seen and no more than one substitution error in formula. A1: Correct equation. Solving their quadratic if full working seen (e.g. use of quadratic equation formula): dM1: At least one correct solution to the quadratic. A1 Showing the two solutions and selecting the correct one. AWRT 1.76. If working not shown in full (e.g. values obtained directly from a calculator): dM1: Obtaining at least one correct solution to the quadratic. A1 Showing the two solutions and selecting the correct one. AWRT 1.76. Note that use of $g = 9.81$ gives the same final answers. | 5 marks

OR

Time Up = 0.9380, Time Down = 0.8221, Total Time = $0.9380 + 0.8221 = 1.76 \text{ s}$ | M2, A1; A1; A1 | M2: Complete method to find total time by adding two (or more) times. A1: Correct time up. A1: Correct time down. A1: Correct total time.

OR

$v^2 = (12 \sin 50°)^2 + 2 \times (-9.8) \times 1$, $v = -8.056$, $-8.056 = 12 \sin 50° - 9.8t$, $t = 1.76s$ | (M1A1), A1; (dM1), (A1) | M1: Using two constant acceleration equations to find $t$. A1: Correct first equation. A1: Seeing AWRT -8.06. dM1: Correct second constant acceleration equation with $\pm$ their $v$. A1 Correct time. Allow AWRT 1.76.

**7. (b)**
$x = 12 \cos 50° \times 1.76 = 13.6 \text{ m}$ | M1A1 | M1: 12cos50° or 12sin50° multiplied by their answer to (a). A1: Correct distance. Accept AWRT 13.6. | 2 marks

**7. (c)**
$13.576 + 3 = 16.576$, $16.576 = V \cos 50°t$, $t = \frac{16.576}{V \cos 50°}$, $1 = V \sin 50°t - 4.9t^2$ OR other vertical equations with vertical kinematics. Other valid working shown. $V = 13.2 \text{ m s}^{-1}$ | M1; dM1, A1F; B1; dM1, A1; A1 | M1: Adding 3 to their answer to part (b). dM1: Horizontal equation with $V \cos 50°$ or $V \sin 50°$ equated to their answer from (b) plus 3. A1F: Correct expression for $t$. Follow through their answer from (b). B1: Correct vertical equation. dM1: Correctly substituting their expression for $t$ into the vertical equation. (Dep on both previous method marks). A1: Correct equation. A1: Correct $V$. Allow AWFW 13.1 and13.3. Condone '16.6' in working. Note that use of $g = 9.81$ gives the same final answer. | 7 marks

OR

$13.576 + 3 = 16.576$, $16.576 = V \cos 50°t$, $V = \frac{16.576}{t \cos 50°}$, $1 = V \sin 50°t - 4.9t^2$, $1 = 16.576 \tan 50° - 4.9t^2$, $t^2 = \frac{16.576 \tan 50° - 1}{4.9}$, $t = 1.9579$, $V = \frac{16.576}{1.9579 \times \cos 50°} = 13.2 \text{ m s}^{-1}$ | M1; dM1, A1F; B1; dM1; A1; A1 | M1: Adding 3 to their answer to part (b). dM1: Horizontal equation with $V \cos 50°$ or $V \sin 50°$ equated to their answer from (b) plus 3. A1F: Correct expression for $V$. Follow through their answer from (b). B1: Correct vertical equation. dM1: Correctly substituting their expression for $V$ into the vertical equation. A1: Correct expression for $t^2$. A1: Correct $V$. Allow AWFW 13.1 and 13.3. Condone '16.6' in working. Note that use of $g = 9.81$ gives the same final answer.

**7. (d)**
Dimensions of ball need to be taken into account. | B1 | B1: Statement about the size of the ball. Ignore irrelevant statements but if weight is zero is included B0. | 1 mark

**Total for Question 7: 15 marks**

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