Answer
(a) The initial velocity is -60.0 m/s
(b) v = -15.5 m/s
Work Step by Step
Let up be the positive direction.
$y(t) = b - ct + dt^2$
$y(t) = (800)~m - (60.0~m/s)~t + (1.05~m/s^2)~t^2$
(a) $v(t) = \frac{dy}{dt} = (-60.0~m/s)+ (2.10~m/s^2)~t$
When t = 0 s,
$v = (-60.0~m/s)+ (2.10~m/s^2)(0) = -60.0~m/s$
The initial velocity of the lander is -60.0 m/s
(b) Let $v$ be the velocity just before landing.
$v^2 = v_0^2+2a(y-y_0)$
$v = \sqrt{v_0^2+2a(y-y_0)}$
$v = \sqrt{(-60.0~m/s)^2+(2)(2.10~m/s^2)(-800~m)}$
$v = -15.5~m/s$
Note that the negative sign means that the lander is descending onto the lunar surface.
