Answer:
[tex]v_r = 42.3 m/s[/tex]
Explanation:
As we know that the width of the river is given as
[tex]w = 30 m[/tex]
now the height is given as
[tex]h = 200 m[/tex]
now by kinematics we have
[tex]h = \frac{1}{2}gt^2[/tex]
[tex]200 = \frac{1}[2}(9.81)t^2[/tex]
[tex]t = 6.38 s[/tex]
now the speed required to cross the river is given as
[tex]v = \frac{w}{t}[/tex]
[tex]v = \frac{30}{6.38}[/tex]
[tex]v = 4.7 m/s[/tex]
now we can use momentum conservation to find the speed of rocket just before collision
[tex]m_r v_r = (m_r + m) v[/tex]
[tex]0.70(v_r) = (5.6 + 0.70)(4.7)[/tex]
[tex]v_r = 42.3 m/s[/tex]
The minimum speed that this rocket must have just before impact is 42.3 m/s.
Given the following data:
Height = 200 m.
Width (distance) = 30 m.
Mass of package = 5.6 kg.
Mass of rocket = 0.70 kg.
The time taken to reach a maximum height is given by this formula:
[tex]H=\frac{1}{2} gt^2\\\\t=\sqrt{\frac{2H}{g} } \\\\t=\sqrt{\frac{2\times 200}{9.8} }[/tex]
t = 6.38 seconds.
For the velocity that is required to cross the river, we have:
[tex]V= \frac{distance}{time} \\\\V=\frac{30}{6.38}[/tex]
V = 4.70 m/s.
Now, we can calculate the minimum speed of the rocket by applying the law of conservation of momentum:
[tex]M_rV_r=(M_r+M)V\\\\0.70V_r=(0.70+5.6)4.70\\\\0.70V_r=29.61\\\\V_r=\frac{29.61}{0.70}[/tex]
Minimum speed = 42.3 m/s.
Read more on speed here: brainly.com/question/10545161
