Answer:
v=6.05 m/s
Explanation:
Given that,
Th initial velocity of the lander, u = 1.2 m/s
The lander is at a height of 1.8 m, d = 1.8 m
We need to find the velocity of the lander at impact. It is a concept based on the conservation of mechanical energy. So,
[tex]\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=W\\\\\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=F\times d\\\\\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=mgd\\\\\dfrac{1}{2}m(v^2-u^2)=mgd\\\\v^2-u^2=2gd[/tex]
v is the velocity of the lander at the impact
g is the acceleration due to gravity on the surface of Mars, which is 0.4 times that on the surface of the Earth, g = 0.4 × 9.8 = 3.92 m/s²
So,
[tex]v=\sqrt{u^2+2gd} \\\\v=\sqrt{(1.2)^2+2\times 9.8\times 1.8} \\\\v=6.05\ m/s[/tex]
So, the velocity of the lander at the impact is 6.05 m/s.