Time Taken by Second Fragment of Split Projectile to Reach Ground

To find the time $t$ taken by the second fragment to reach the ground after the projectile splits, we analyze the motion before and after the split step-by-step.

1. Motion before the split

The projectile is launched from origin $O$ with an initial velocity $u = 5\sqrt{2} \text{ m/s}$ at an angle of $45^\circ$ with the vertical (which is also $45^\circ$ with the horizontal).

The horizontal and vertical components of the initial velocity are: $u_x = u \cos 45^\circ = 5\sqrt{2} \times \frac{1}{\sqrt{2}} = 5 \text{ m/s}$ $u_y = u \sin 45^\circ = 5\sqrt{2} \times \frac{1}{\sqrt{2}} = 5 \text{ m/s}$

The time taken to reach the maximum height (peak of trajectory) is: $t_h = \frac{u_y}{g} = \frac{5}{10} = 0.5 \text{ s}$

The maximum height $H$ reached by the projectile is: $H = \frac{u_y^2}{2g} = \frac{5^2}{2 \times 10} = 1.25 \text{ m}$

At the highest point, the vertical component of velocity is zero, so the velocity vector of the projectile just before the split is: $\vec{v} = 5\hat{i} \text{ m/s}$

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2. Conservation of Linear Momentum during the Split

Let the mass of the projectile be $m$. It breaks into two equal parts, each of mass $\frac{m}{2}$. By the law of conservation of linear momentum: $m \vec{v} = \frac{m}{2} \vec{v}_1 + \frac{m}{2} \vec{v}_2$ $\vec{v}_1 + \vec{v}_2 = 2\vec{v} = 10\hat{i} \text{ m/s}$

In component form:

1. $v_{1x} + v_{2x} = 10 \text{ m/s}$
2. $v_{1y} + v_{2y} = 0 \text{ m/s}$

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3. Motion of the First Fragment

The first part falls vertically downwards, which means its horizontal velocity is zero: $v_{1x} = 0 \implies v_{2x} = 10 \text{ m/s}$

We are given that the first fragment strikes the ground $0.5 \text{ s}$ after splitting. Using the equation of motion for vertical displacement ($h = 1.25 \text{ m}$): $h = v_{1y} t_1 + \frac{1}{2} g t_1^2$ $1.25 = v_{1y}(0.5) + \frac{1}{2}(10)(0.5)^2$ $1.25 = 0.5 v_{1y} + 1.25 \implies v_{1y} = 0 \text{ m/s}$

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4. Motion of the Second Fragment

Since $v_{1y} = 0 \text{ m/s}$, the vertical velocity component of the second fragment immediately after splitting is: $v_{2y} = 0 \text{ m/s}$

Since it is at the same initial height $H = 1.25 \text{ m}$ with zero initial vertical velocity, the time $t$ it takes to reach the ground is determined purely by free fall under gravity: $H = v_{2y} t + \frac{1}{2} g t^2$ $1.25 = 0 + \frac{1}{2} (10) t^2$ $5 t^2 = 1.25 \implies t^2 = 0.25 \implies t = 0.5 \text{ s}$

Thus, the value of $t$ is $0.5$.