What is the Rover's landing speed? (Picture)

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1 Answer
Mar 13, 2018

4.65 ms^-1

Explanation:

Total energy remains the same during this motion,as motion is only due to the effect of gravity.

So,total energy on top of crater was m(g/3)h +1/2 mv^2=m(g/3)*1.95+1/2 m 3^2(Potential energy +Kinetic energy)

While landing it will have purely kinetic energy,so at that time if velocity is v then,total energy =1/2 mv^2

So,(1.95mg)/3 +9/2 m=1/2 mv^2

Solving we get, v=4.65 ms^-1

ALTERNATIVELY

The Mars rover will follow a parabolic path in its journey from the 1.95m tall crater to the region infront of the bottom of the crater.

Suppose,while landing it will have vertical component of velocity v_y,so considering vertical motion only we can write, v_y^2 =0^2 +2(g/3)*1.95 using v^2=u^2+2as the rover was launched with an initial velocity horizontally,so initially it had no vertical component of velocity and a=g/3

So,v_y^2=12.74

Now,horizontal component of velocity remains the same i.e v_x=3 m/s

So,landing velocity = sqrt(v_x^2 +v_y^2)=3^2 +12.74=4.65ms^-1
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