Question

A lunar lander is descending toward the moon's surface. Until the lander reaches the surface, its height above the surface of the moon is given by

y(t)=b−ct+dt^2 , where b = 710m is the initial height of the lander above the surface, c = 64.0m/s, and d = 1.03 m/s^2.

(A) What is the initial velocity of the lander, at t = 0?

(B) What is the velocity of the lander just before it reaches the lunar surface?

Answer 1

The time by when the lunar lander lands the surface is found by solving y(t)=0.

b - ct + d t² = 0 which implies t = (c +- ( $\sqrt{c^2 - 4 b d}$ )) / (2d)
t = (64 m/s +- ( $\sqrt{(64m/s)^2 - 4 * 710 m * 1.03 m/s^2}$ ))/(2*1.03 m/s²)
t = (64.0 m/s +- 34.21 m/s) / (2.6 m/s²)
t = 11.53 s or t = 37.69 s .

Obviously we need the smallest time solution (when y equals zero from the first time, the second solution being non-physical as it is the solution corresponding to the lunar lander going down through the surface and coming back up to y=0 from under the surface)

So y=0 when t=11.53 s

The velocity can now be found from:

v(t) = dy/dt = -c + 2 d t

So at the surface we have

v(11.53s) = -64.0 m/s + 2 * 1.03 m/s² * 11.53s
= -40.24 m/s

a downward velocity of 40.24 m/s