Question 3.2.2: Additional Answer Options

2021 - November - Paper 1 - Question 3

Or

Option 2

Upwards as positive:

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

-200=(2,96)\Delta t+1/2(-9,8)\Delta t^2

\Delta t=6,70\mathrm{s}(6,697)

Or

Downwards as positive:

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

200=(-2,96)\Delta t+1/2(9,8)\Delta t^2

\Delta\mathrm{t}=6,70\mathrm{s}\quad(6,697)

Or

Option 3

Upwards as positive:

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

-200=\left(\frac{+2,96+(-62,68)}{2}\right)\Delta t

\Delta\mathrm{t}=6,70\mathrm{s}(6,698)

Or

Downwards as positive:

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

200=\left(\frac{-2,96+62,68}{2}\right)\Delta t

\Delta\mathrm{t}=6,70\mathrm{s}(6,698)

Or

Option 4

Upwards as positive:

From 200 m upwards:

v_{f}=v_{i}+a\Delta t

0=2,96+(-9,8)\Delta t

\Delta t=0,3\mathrm{~s}(0,302)

From max h downwards:

v_{f}=v_{i}+a\Delta t

-62,68=0+(-9,8)\Delta t

\Delta\mathrm{t}=6,40\mathrm{~s}(6,369)

t_{A}=0,3+6,40=6,7\mathrm{~s}

Or

Downwards as positive:

From 200 m upwards:

\mathrm{v}=\mathrm{v}_{\mathrm{i}}+\mathrm{a}\Delta\mathrm{t}

0=-2,96+(9,8)\Delta t

\Delta t=0,3\mathrm{~s}(0,302)

From max h downwards:

v_{f}=v_{i}+a\Delta t

62,68=0+(9,8)\Delta t

\Delta\mathrm{t}=6,40\mathrm{~s}(6,369)

t_{A}=0,3+6,40=6,7\mathrm{~s}

Or

Option 5

Upwards as positive:

From 200 m upwards:

v_{f}=v_{i}+a\Delta t

0=2.96+(-9.8)\Delta t

\Delta t=0,3\mathrm{~s}(0,302)

From max h downwards:

v_{f}=v_{i}+a\Delta t

-62,68=-2,96+(-9,8)\Delta t

\Delta\mathrm{t}=6,09\mathrm{~s}(6,094)

t_{A}=2(0,3)+6,09=6,69s

Or

Downwards as positive:

From 200 m upwards:

v_{f}=v_{i}+a\Delta t

0=-2,96+(9,8)\Delta t

\Delta\mathrm{t}=0,3\mathrm{~s}(0,302)

From max h downwards:

\mathrm{v}_{\mathrm{f}}=\mathrm{v}_{\mathrm{i}}+\mathrm{a}\Delta\mathrm{t}

62,68=2,96+(9,8)\Delta t

\Delta t=6,09\mathrm{~s}(6,094)

t_{A}=2(0,3)+6,09=6,69\mathrm{~s}

Or

Option 6

Upwards as positive:

F_{net}\Delta t=m\left(v_{f}-v_{i}\right)

mg\Delta t=m\left(v_{f}-v_{i}\right)

g\Delta t=v_{f}-v_{i}

(-9,8)\Delta t=(-62,68)-(2,96)

\Delta\mathrm{t}=6,69\mathrm{~s}

Or

Downwards as positive:

F_{net}\Delta t=m\left(v_{f}-v_{i}\right)

mg\Delta t=m\left(v_{f}-v_{i}\right)

g\Delta t=v_{f}-v_{i}

(9,8)\Delta t=62,68-(-2,96)

\Delta\mathrm{t}=6,69\mathrm{~s}

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