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+dt2, where b = 800 m is the initial height of the lander above the surface, c = 63.0 m/s , and d = 1.05 m/s2What is the initial velocity of the lander, at t = 0 and just before it reaches the lunar surface?

Answer 1

Answer:

Part A:

$v_y=-63 m/s$

Part B:

$v_y=-24.67792 m/s$

Explanation:

Part A:

The formula we are going to use is:

$v_{y}= \frac{dy(t)}{dt} \\v_y=\frac{d(b-ct+dt^2)}{dt} \\v_y=-c+2d*t$

For finding the initial velocity we put t=0 in above formula:

$v_y=-c+2d*t$

$v_y=-c+2d*0$

$v_y=-c$

c=63 So,

$v_y=-63 m/s$

Part B:

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

$y(t)=b-ct+dt^2=0$

Using Quadratic formula:

$t=\frac{-b\±\sqrt{b^2-4ac} }{2a}$

In our case:

$t=\frac{c\±\sqrt{c^2-4db} }{2d}$

$t=\frac{63\±\sqrt{(63)^2-4*1.05*800} }{2*1.05}$

$t=41.75139sec$                $t=18.248606sec$

We are going to choose t=18.248606 sec because it is smaller.

$v_y=-c+2d*t$ (Calculated above)

$v_y=-63+2(1.05)*18.248606$

$v_y=-24.67792 m/s$