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# A girl after being angry throws her engagement ring from the top of a building $12\,{\text{m}}$ high towards her boyfriend with an initial horizontal speed of $5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$, speed with which the ring touches the ground is:A. $5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$B. $14.3\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$C. $1.5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$D. $16.2\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$

Last updated date: 20th Jun 2024
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Hint:Use the expression for the kinematic equation for final vertical velocity in terms of the vertical displacement of the object. Using this kinematic equation, calculate the vertical component of the speed with which the ring touches the ground and then calculate the net speed from horizontal and vertical components of the speed.

Formula used:
The kinematic equation for the final vertical velocity ${v_y}$ in terms of vertical displacement is
$v_y^2 = u_y^2 + 2gh$ …… (1)
Here, ${u_y}$ is the initial vertical velocity of the object, $g$ is acceleration due to gravity and $h$ is the vertical displacement of the object.

We have given that the height of the building from the ground is $12\,{\text{m}}$.
$h = 12\,{\text{m}}$
We have also given that the horizontal component of the speed with which the girl throws her engagement ring is $5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$.
${u_x} = 5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
This horizontal component of the initial speed is the same when the ring touches the ground.
${v_x} = 5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$

We have asked to calculate the speed with which the ring touches the ground.Let us first calculate the vertical component of the speed with which the ring touches the ground using equation (1).The vertical component of the initial vertical speed of the ring is zero as it has only horizontal component initially.
${u_y} = 0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
Substitute $0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$ for ${u_y}$, $10\,{\text{m/}}{{\text{s}}^2}$ for $g$ and $12\,{\text{m}}$ for $h$ in equation (1).
$v_y^2 = {\left( {0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)^2} + 2\left( {10\,{\text{m/}}{{\text{s}}^2}} \right)\left( {12\,{\text{m}}} \right)$
$\Rightarrow v_y^2 = 2\left( {10\,{\text{m/}}{{\text{s}}^2}} \right)\left( {12\,{\text{m}}} \right)$
$\Rightarrow v_y^2 = 240$
$\Rightarrow {v_y} = \sqrt {240} \,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
Hence, the vertical component of the speed with which the ring touches the ground is $\sqrt {240} \,{\text{m}} \cdot {{\text{s}}^{ - 1}}$.

The net speed with which the girl throws the ring is given by
$v = \sqrt {v_x^2 + v_y^2}$
Substitute $5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$ for ${v_x}$ and $\sqrt {240} \,{\text{m}} \cdot {{\text{s}}^{ - 1}}$ for ${v_y}$ in the above equation.
$v = \sqrt {{{\left( {5\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)}^2} + {{\left( {\sqrt {240} \,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)}^2}}$
$\Rightarrow v = \sqrt {25 + 240}$
$\Rightarrow v = \sqrt {265}$
$\therefore v = 16.2\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
Therefore, the speed with which the ring touches the ground is $16.2\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$.

Hence, the correct option is D.

Note: The students should keep in mind that we have given a horizontal component of the speed with which the girl throws the ring. Since the ring is in free fall when the girl throws it, the horizontal component of speed of the ring remains the same and there is only change in vertical component of the speed. Hence, we have used the same initial horizontal component of speed as the final horizontal component of the speed.