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A stone is dropped from a height h simul...

A stone is dropped from a height h simultaneously another stone is thrown up from the ground with such a velocity so that it can reach a height of 4h. The time when two stones cross each other is `sqrt(((h)/(kg)))` where `k = "_____"`

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To solve the problem step by step, we will analyze the motion of both stones and find the time at which they cross each other. ### Step 1: Understand the scenario - A stone is dropped from a height \( h \) (let's call this Stone A). - Another stone is thrown upwards from the ground with enough velocity to reach a maximum height of \( 4h \) (let's call this Stone B). ### Step 2: Determine the initial velocity of Stone B To find the initial velocity \( U \) of Stone B, we can use the kinematic equation for maximum height: \[ v^2 = u^2 + 2as \] At the maximum height, the final velocity \( v = 0 \), the acceleration \( a = -g \) (since it is moving against gravity), and the displacement \( s = 4h \). Thus, we have: \[ 0 = U^2 - 2g(4h) \] Rearranging gives: \[ U^2 = 8gh \] So, the initial velocity \( U \) is: \[ U = \sqrt{8gh} = 2\sqrt{2gh} \] ### Step 3: Analyze the motion of both stones - **For Stone A (dropped from height \( h \))**: - Initial velocity \( u_A = 0 \) - Displacement \( d_A = \frac{1}{2} g t^2 \) - **For Stone B (thrown upwards)**: - Initial velocity \( u_B = U = 2\sqrt{2gh} \) - Displacement \( d_B = U t - \frac{1}{2} g t^2 \) ### Step 4: Set up the equation for when the stones cross each other At the time \( t \) when the two stones cross each other, the sum of their displacements must equal the height \( h \): \[ d_A + d_B = h \] Substituting the expressions for \( d_A \) and \( d_B \): \[ \frac{1}{2} g t^2 + \left(2\sqrt{2gh} t - \frac{1}{2} g t^2\right) = h \] This simplifies to: \[ 2\sqrt{2gh} t = h \] ### Step 5: Solve for time \( t \) Rearranging gives: \[ t = \frac{h}{2\sqrt{2gh}} = \frac{h}{2\sqrt{2g}\sqrt{h}} = \frac{\sqrt{h}}{2\sqrt{2g}} \] ### Step 6: Express in the required form The expression we derived for \( t \) is: \[ t = \frac{\sqrt{h}}{2\sqrt{2g}} = \sqrt{\frac{h}{8g}} \] This matches the form given in the question \( \sqrt{\frac{h}{kg}} \), where \( k = 8 \). ### Conclusion Thus, the value of \( k \) is: \[ \boxed{8} \]

To solve the problem step by step, we will analyze the motion of both stones and find the time at which they cross each other. ### Step 1: Understand the scenario - A stone is dropped from a height \( h \) (let's call this Stone A). - Another stone is thrown upwards from the ground with enough velocity to reach a maximum height of \( 4h \) (let's call this Stone B). ### Step 2: Determine the initial velocity of Stone B To find the initial velocity \( U \) of Stone B, we can use the kinematic equation for maximum height: ...
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