Answer:

Volume of the mass after 5 seconds (at the time of explosion)

v=u-g+= 1cm -(16) (5)

= 50ms upwards

Since no external force acts on the system in the explosion, the linear momentum of the system remains conserved .

m v=m¹ v¹ + m² v²

1 (50)=0.4 (-25)+0.6 v²

v²= 100ms

Hence The After project from ground move 100ms.

Explanation:

Please Mark me Brainliest

Answer:

The problem involves projectile motion, and we can use the principles of conservation of energy to solve it. The change in momentum is related to the change in kinetic energy.

Given:

- Mass of the ball (\(m\)) = 0.1 kg

- Change in momentum (\(\Delta p\)) = 1 kg·m/s

- Range is twice the maximum height (\(R = 2H\))

The change in kinetic energy (\(\Delta KE\)) is given by:

\[ \Delta KE = \frac{1}{2} m v^2 - \left(\frac{1}{2} m u^2\right) \]

where:

- \(v\) is the final velocity at the highest point (which is 0 m/s since the ball momentarily stops at the highest point),

- \(u\) is the initial velocity (the speed of projection).

Given that \(\Delta p = m \cdot \Delta v\), where \(\Delta v\) is the change in velocity, we have:

\[ \Delta v = v - u \]

Substituting this into the expression for \(\Delta KE\), we get:

\[ \Delta KE = \frac{1}{2} m v^2 - \left(\frac{1}{2} m (u + \Delta v)^2\right) \]

Now, we can use the fact that \(R = \frac{u^2 \sin(2\theta)}{g}\), where \(\theta\) is the angle of projection and \(g\) is the acceleration due to gravity.

For maximum height (\(H\)), the vertical component of the velocity is \(0\). So, \(v_y = 0\) at the highest point. We have:

\[ v_y^2 = u^2 \sin^2(\theta) - 2gH \]

\[ 0 = u^2 \sin^2(\theta) - 2gH \]

Solving for \(H\):

\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

Now, substitute \(H\) into the expression for \(R\):

\[ R = \frac{u^2 \sin(2\theta)}{g} \]

\[ R = \frac{2u^2 \sin(\theta)\cos(\theta)}{g} \]

\[ R = \frac{u^2 \sin(2\theta)}{g} \]

Now, we know that \(R = 2H\), so:

\[ \frac{u^2 \sin(2\theta)}{g} = 2 \cdot \frac{u^2 \sin^2(\theta)}{2g} \]

\[ \sin(2\theta) = 2 \sin^2(\theta) \]

\[ \sin(2\theta) = 2(1 - \cos^2(\theta)) \]

\[ \sin(2\theta) = 2 - 2\cos^2(\theta) \]

\[ \cos^2(\theta) = \frac{1}{2} \]

\[ \cos(\theta) = \frac{1}{\sqrt{2}} \]

\[ \cos(\theta) = \frac{\sqrt{2}}{2} \]

\[ \theta = \frac{\pi}{4} \]

Now, substitute \(u\) and \(g\) into the equation for \(R\):

\[ R = \frac{u^2 \sin(\frac{\pi}{2})}{g} \]

\[ R = \frac{u^2}{g} \]

\[ R = \frac{u^2}{9.8} \]

Now, substitute \(R = 2H\):

\[ \frac{u^2}{9.8} = 2 \cdot \frac{u^2 \sin^2(\frac{\pi}{4})}{2 \cdot 9.8} \]

\[ \frac{u^2}{9.8} = \frac{u^2}{9.8} \]

This equation is satisfied, so the given conditions are consistent with the projectile motion. Now, since the projectile motion is consistent, we can use the initial velocity \(u\) to find the speed of projection.

\[ \Delta KE = \frac{1}{2} m v^2 - \left(\frac{1}{2} m u^2\right) \]

\[ 1 = \frac{1}{2} \cdot 0 - \left(\frac{1}{2} \cdot (u^2)\right) \]

\[ 1 = -\frac{1}{2} u^2 \]

\[ u^2 = -2 \]

\[ u = \sqrt{-2} \] (ignoring the negative square root because speed is non-negative)

\[ u = i\sqrt{2} \]

So, the correct option is not among the provided choices. There might be an issue with the given answer choices or the problem statement. Please double-check the options or the problem statement for accuracy.

Explanation: