Question 3.2.1: Additional Answer Options

2022 - May/June - Paper 1 - Question 3

Or

Option 2

Upwards as positive:

v_{f}=v_{i}+a\Delta t(1)

-15=0+(-9,8)\Delta t (2)

\Delta t=1,53\mathrm{~s} (3)

Or

Downwards as positive:

v_{f}=v_{i}+a\Delta t (1)

15=0+(9,8)\Delta t (2)

\Delta t=1,53\mathrm{~s} (3)

Or

Option 3

Upwards as positive:

v_{f}=v_{i}+a\Delta t (1)

-15=15+(-9,8)\Delta t (2)

\Delta t=3,06\mathrm{~s}

\Delta t=1,53\mathrm{~s} (3)

Or

Downwards as positive:

v_{f}=v_{i}+a\Delta t (1)

15=-15+(9,8)\Delta t (2)

\Delta t=3,06\mathrm{~s}

\Delta t=1,53\mathrm{~s} (3)

Or

Option 4

Upwards as positive:

\Delta y=v_{i}\Delta t+1/2a\Delta t^2 (1)

0=(15)\Delta t+1/2(-9,8)\Delta t^2 (2)

\Delta t=3,06\mathrm{~s}

\Delta t=1,53\mathrm{~s} (3)

Or

Downwards as positive:

\Delta y=v_{i}\Delta t+1/2a\Delta t^2 (1)

0=(-15)\Delta t+1/2(9,8)\Delta t^2 (2)

\Delta t=3,06\mathrm{~s}

\Delta\mathrm{t}up=1,53\mathrm{~s} (3)

Or

Option 5

Part 1

\left(\mathrm{E}_{\text{mech }}\right)_{\text{Top }}=\left(E_{\text{mech }}\right)_{30m}

\left(E_{P}+E_{K}\right)_{Top}=\left(E_{P}+E_{K}\right)_{30m}

\left(mgh+1/2mv^2\right)_{\text{Top }}=\left(mgh+1/2mv^2\right)_{30m}

(9,8)h+0=0+(1/2)(15)^2

\Delta\mathrm{h}=11,48\mathrm{~m}

Or

W_{nc}=\Delta K+\Delta U

W_{nc}=\Delta K+mg\left(h_{f}-h_{i}\right)

0=1/2mv_{f}^2-1/2mv_{i}^2+mgh_{f}-mgh_{i}

0=1/2\left(0-15^2\right)+(9,8)\Delta h

\Delta\mathrm{h}=11,48\mathrm{~m}

Or

W_{\text{net }}=\Delta E_{k}

w\Delta y\cos\theta=1/2mv_{f}^2-1/2mv_{i}^2

(9,8)\Delta y\cos180^{\circ}=0-1/2(15)^2

\Delta y=11,48m

Part 2

\Delta\mathrm{y}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2 (1)

11,48=(15)\Delta t+1/2(-9,8)\Delta t^2 (2)

\Delta\mathrm{t}=1,53\mathrm{~s} (3)

Or

\Delta y=v_{i}\Delta t+1/2a\Delta t^2 (1)

11,48=(-15)\Delta t+1/2(9,8)\Delta t^2 (2)

\Delta\mathrm{t}=1,53\mathrm{~s} (3)

Or

\Delta y=\left(\frac{v_{i}+v_{f}}{2}\right)\Delta t (1)

{11,48}=\left(\frac{15+0}{2}\right)\Delta t (2)

\Delta t=1,53\mathrm{~s} (3)

Or

Option 6

Upwards as positive:

\Delta\mathrm{y}=\left(\frac{\mathrm{v}_{\mathrm{i}}+\mathrm{v}_{\mathrm{f}}}{2}\right)\Delta\mathrm{t} (1)

\Delta\mathrm{y}=\left(\frac{15+0}{2}\right)\Delta\mathrm{t}

\Delta y=7,5\Delta t

v_{f}^2=v_{i}^2+2a\Delta y

0=(15)^2+2(-9,8)(7,5\Delta t) (2)

\Delta\mathrm{t}=1,53\mathrm{~s} (3)

Or

Downwards as positive:

\Delta y=\left(\frac{v_{i}+v_{f}}{2}\right)\Delta t

\Delta\mathrm{y}=\left(\frac{-15+0}{2}\right)\Delta\mathrm{t}

\Delta y=-7,5\Delta t

v_{f}^2=v_{i}^2+2a\Delta y (1)

0=(-15)^2+2(9,8)(-7,5\Delta t) (2)

\Delta t=1,53\mathrm{~s} (3)

Or

Option 7

Upwards as positive:

\left.\begin{array}{l}F_{net}\Delta t=m\Delta v\\ F_{net}\Delta t=m\left(v_{f}-v_{i}\right)\end{array}\right\}Any one (1)

-(9,8)\Delta t=0-15 (2)

\Delta t=1,53\mathrm{~s} (3)

Or

Downwards as positive:

\left.\begin{array}{l}F_{net}\Delta t=m\Delta v\\ F_{net}\Delta t=m\left(v_{f}-v_{i}\right)\end{array}\right\}Any one (1)

(9,8)\Delta t=15-0 (2)

\Delta t=1,53\mathrm{~s} (3)

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