A $4 k g$ projectile is launched vertically from a height of $2 m$ with an initial speed of $12 m s^{- 1}$ . Find its speed (to nearest integer) when it is at a height of $3 m$ above the ground.

10 m s^{- 1}

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11 m s^{- 1}

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12 m s^{- 1}

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13 m s^{- 1}

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Solution

The correct option is B $11 m s^{- 1}$
Total mechanical energy of projectile remains constant.
$E_{o} = E_{f}$
Mechanical energy consists of kinetic and potential energy.
$U_{o} + K_{o} = U_{f} + K_{f}$
Using formulas of kinetic and potential energy:
$m g h_{o} + \frac{1}{2} m v_{o}^{2} = m g h_{f} + \frac{1}{2} m v_{f}^{2}$
$v_{f} = \sqrt{2 g (h_{o} - h_{f}) + v_{o}^{2}}$
Subsituting values, we get
$v_{f} = \sqrt{2 \times 10 \times (2 - 3) + 12^{2}}$
$v_{f} = \sqrt{144 - 20} = \sqrt{124}$
$v_{f} \approx 11.14 m s^{- 1}$
Rounding to nearest integer, final speed is 11 m/s.

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A 4 kg projectile is launched vertically from a height of 2 m with an initial speed of 12 ms−1. Find its speed (to nearest integer) when it is at a height of 3 m above the ground.

A $4 k g$ projectile is launched vertically from a height of $2 m$ with an initial speed of $12 m s^{- 1}$ . Find its speed (to nearest integer) when it is at a height of $3 m$ above the ground.