A lunar lander is descending toward the moon\'s surface. Until the lander reache

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)=bct+dt2, where b = 710 m is the initial height of the lander above the surface, c = 65.0 m/s , and d = 1.00 m/s2 . Part A What is the initial velocity of the lander, at t = 0? Express your answer with the appropriate units. Part B What is the velocity of the lander just before it reaches the lunar surface? Express your answer with the appropriate units.

Explanation / Answer

Subsing the given values into the given equation for height above the lunar surface:

y(t) = 710 - 65.0t + 1.0t²

when the lander reaches the lunar surface y(t) = 0

so 1.0t² - 65.0t + 710 = 0

t = [-b ± (b² - 4ac)] / 2a

t = [65.0 ± (65.0² - 4 * 1.0 * 710)] / 2.0

t = [65.0 ± 1385] / 2.0

t = 51.10 or t =13.895

Now the parabola opens upwards b/c the coefficient of the t² term is positive

so the lander reaches the lunar surface at t = 13.895 ... [reject t = 51.10 b/c that is the time where the parabola crosses the t-axis again after being below the t-axis and the lander doesn't go below the lunar surface so it doesn't go below the t-axis

all the above can be left out if you use the second method shown for finding the velocity just b4 touchdown



v(t) = y '(t) = 2.0t - 65.0

when t = 0, v(t) = 2.0 * 0 - 65.0 = -65.0 m/s ... [the lander is moving down and its velocity is negative so UP has been taken as the positive direction]

so the initial velocity of the lander is 65.0 m/s downwards (-65.0 m/s)




Method 1 for finding velocity just prior to touchdown:

It takes the lander 13.895 s to reach the surface

so when t = 13.895, v(t) = 2.0 * 27.79 - 65.0 = -9.42 m/s

so the velocity of the lander just b4 it reaches the lunar surface is 9.42m/s downwards (-9.42 m/s)


Method 2 for finding velocity just prior to touchdown:

This method is probably a quicker way to go b/c if you do it this way you don't have to solve the quadratic at the start so all that can be left out and just do this:

a(t) = v '(t) = 2.0 m/s²

so the acceleration of the lander is constant for the descent and is 2.0 m/s² upwards (b/c it's positive)

v(f)² = v(i)² + 2as

v(f) = [v(i)² + 2as]

v(i) = -65.0 m/s

a = 2.0 m/s²

s = -710 m ... [displacement is down ... so negative]

v(f) = [(-65.0)² + 2 * 2.0 * (-710)]

v(f) = -9.42 m/s ... [taking the negative result b/c the lander is moving downwards]

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