Given:

“M = 1800~\text{kg};~a = 85.0~\frac{\text{m}}{\text{s}^2};~t = 2.5~\text{s}; \\nm_1 = 2m_2;~{\alpha}_2=10^\circ;~h_2 = 9000~\text{m}. \\”

First of all, let’s note that we can determine the height and speed (naturally upwards vertical) of the intact rocket just at the moment of the explosion as:

“h = \frac{1}{2}at^2;~v = at.~”

Let’s also remember that we can determine the speed of the smaller piece (just after the explosion), making use of the equation for height of its ballistic trajectory:

“h_2 = h + \frac{{v_2}^2{\cos^2{\alpha}_2}}{2g}; n\implies v_2 = \frac{\sqrt{g(2h_2-at^2)}}{\cos\alpha_2}.”

a)

Determine the speed and direction of the larger, two times as massive, piece?

We’ll assume that the larger piece directed at the east-upwards, at an angle “{\alpha}_1~”between its direction and the upwards vertical (original rocket direction), at a speed “v_1~”immediately after the explosion.

Let’s use the conservation of momentum just before and after the explosion:

“\begin{cases}n0 = m_1v_1\sin{\alpha}_1 – m_2v_2\sin{\alpha}_2; \\nMv = m_2v_2\cos{\alpha}_2 + m_1v_1\cos{\alpha}_1.n\end{cases}”

Solving it for the unknowns,

“\tan\alpha_1 = \frac{m_2v_2\sin\alpha_2}{m_2v_2\cos\alpha_2-Mv} = \tan\alpha_2\frac{m_2\sqrt{g(2h_2-at^2)}}{m_2\sqrt{g(2h_2-at^2)}-Mat}; \\”

“v_1 = \frac{1}{m_1}\sqrt{\frac{m_2^2g(2h_2-at^2)}{\cos^2\alpha_2} + (Mat)^2 – 2Mm_2at\sqrt{g(2h_2-at^2)}}; \\”

We can simplify those a bit by canceling masses if we recall that “m_1 = 2m_2; M = 3m_2.”

“v_1 = \frac{1}{2}\sqrt{\frac{g(2h_2-at^2)}{\cos^2\alpha_2} + (3at)^2 – 6at\sqrt{g(2h_2-at^2)}}; \\”

“\alpha_1 = \arctan{\lbrack\tan(\alpha_2)\frac{\sqrt{g(2h_2-at^2)}}{3at-\sqrt{g(2h_2-at^2)}}\rbrack}; \\”

With numerical values,

“v_1 \approx 117.57~\frac{\text{m}}{\text{s}}; \\n\alpha_1 \approx 18.08^\circ. \\”

b)

In order to find the horizontal range of the larger piece we will have to write its kinematic equations for horizontal and vertical axes, using the time reference “\tau~” from the moment of the explosion:

“\begin{cases}nx_1 = v_1\sin\alpha_1\tau; \\n0 = h + v_1\cos\alpha_1\tau – \frac{1}{2}g\tau^2. \\n\end{cases}”

The last equation shows the height 0 at the moment of eventual falling.

Rewriting it in form of quadratic equation and replacing h, we get and solve:

“g\tau^2 – 2v_1\cos{(\alpha_1)}\tau – at^2 = 0; \\n\tau_{i,j} = \frac{v_1\cos\alpha_1 \pm \sqrt{v_1^2\cos^2\alpha_1 + gat^2}}{g}; \\”

Obviously we need the positive solution as the negative one corresponds to the wrong direction of time and, correspondingly, direction of movement.

As the formulas would become too cumbersome if were to express the values through the given original values, we will use some intermediate calculations:

“\tau \approx 24.96~\text{s}; \\nx_1 = 117.57~\frac{\text{m}}{\text{s}}~*~\sin{(18.08^\circ)}~*~24.96~\text{s} \approx 910.82~\text{m}.”

The rescue team should look in east direction from the launch site.