Answer:
v=6.05 m/s
Step-by-step 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,
![(1)/(2)mv^2-(1)/(2)mu^2=W\\\\(1)/(2)mv^2-(1)/(2)mu^2=F* d\\\\(1)/(2)mv^2-(1)/(2)mu^2=mgd\\\\(1)/(2)m(v^2-u^2)=mgd\\\\v^2-u^2=2gd](https://img.qammunity.org/2021/formulas/physics/college/ysbw5ilq3axf5mkx2069zeolh9t4rk65qi.png)
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,
![v=√(u^2+2gd) \\\\v=√((1.2)^2+2* 9.8* 1.8) \\\\v=6.05\ m/s](https://img.qammunity.org/2021/formulas/physics/college/fgh4c75m94o7unjdztl8ozkz5nzcqkeutb.png)
So, the velocity of the lander at the impact is 6.05 m/s.