Solution to Test Paper 2

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Solutions
1 (a) \displaystyle \begin{align}

& 44 \ \textrm{.}1=\frac{1}{2}\times 9 \ \ 8{{t}^{2}} \\ & t=\sqrt{\frac{44 \textrm{.}1}{4 \textrm{.}9}}=3\text{ s} \\ \end{align}

OR

\displaystyle \begin{align} & s=\frac{1}{2}\times 9 \textrm{.}8\times {{3}^{2}}=44 \textrm{.}1 \\ & \text{AG} \\ & \therefore \text{Hits ground after 3 seconds} \\ \end{align}



M1

A1

A1


(M1)

(A1)

(A1)


(3 marks) M1: Use of constant acceleration

equation with \displaystyle v=0

A1: Correct equation

A1: Correct \displaystyle s


1 (b) \displaystyle \begin{align}

& {{v}^{2}}={{0}^{2}}+2\times 9 \textrm{.}8\times 44 \textrm{.}1 \\ & v=\sqrt{864 \textrm{.}36}=29 \ \textrm{.}4\text{ m}{{\text{s}}^{-1}} \\ \end{align}

OR

\displaystyle \begin{align} & v=0+9 \textrm{.}8\times 3 \\ & v=29 \textrm{.}4 \text{ m}{{\text{s}}^{-1}} \\ \end{align}


M1

A1

A1


(3 marks)


M1: Use of constant acceleration equation with \displaystyle v=0

A1: Correct equation.

A1: Correct \displaystyle v.


1 (c) Air resistance would slow the ball down. B1 (1 mark) B1: Sensible statement about air resistance.
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