Suppose an astronaut jumps from the bottom of a ladder on the moon.

Question

It is 24 cm from the ground when it jumps. a gyroscope can accurately detect when the speed of the astronauts is greater than 1.0 m/s. the gyroscope detects his velocity to be 1.0 m/s (downwards) 2.80 s after he left the ladder. the acceleration due to gravity on the moon is 1.6 m/s^2

A) velocity and position as a function of time above the surface of the moon, including the initial velocity

B) The maximum height of the astronaut and at what time this occurs

C)The time of the astronaut just before he touches the moon

Answer 1

height = s(t) =-1.6t^2 +8.96 + 24

s(t) =1.6t^2 + vot +24 where vo = initial velocity = 8.96

hits the ground when s=0

v(2.8) = -1 foot per second at time t=2.8 seconds

s'=v = -3.2t+vo

s'(2.8) = -3.2(2.8)+vo=-1

vo = -1+9.96

vo =8.96 feet per second = initial velocity

s(t) = -1.6t^2 +8.96t +24 = height at time t

v(t)=s'(t) =032t +8.96 = velocity at time t

s(t) = -1.6t^2 +8.96+24 = 0

= t^2 - 5.6t - 15=0

t^2 -5.6 + 2.8^1 = 15+2.8^2

(t-2.8)^2 =22.84

t-2.8 = 2sqr5.71

t = 2.8+2sqr5.71= about 2.8+4.78

t= about 7.58 seconds to reach zero h, when he just touches the moon

max height is when s' = 0

s' = -3.2t + 8.96 = 0

t = 8.98/3.2 = 2.8 seconds to reach maximum height

max height = s(2.8)= -1.6(2.8)^2 +8.96(2.8) + 24

= 36.544 feet