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
$11 m s^{- 1}$

Substituting the values of mass, height and velocity of the shot in the formula:
Total energy = Potential energy + Kinetic energy = $m g h + \frac{1}{2}$ m $v^{2}$
Total energy = $4 (10 \times 2 + 0.5 \times$ 12 $\times 12) = 367 J$
Now, Potential energy = $m g h$ = $4 \times 10 \times 3 = 120 J$
Kinetic energy = Total energy – Potential energy = $247 J$
$K . E . = \frac{1}{2}$ $m v^{2}$
$v = \sqrt{\frac{2 K E}{m}}$ = $\sqrt{\frac{2 \times 247}{4}}$ = $11 m s^{- 1}$ .

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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.