Observing a spacecraft land on a distant asteroid, scientists notice that the craft is falling at a rate of 4 m/s. when it is 180 m closer to the surface of the asteroid, the craft reports a velocity of 9 m/s. according to their data, what is the approximate gravitational acceleration on this asteroid?

0.18 m/s^2  
Let's use the definition of energy and work to solve this problem. To make the math simple, I'll use a mass of 2 kg for the spacecraft. 
Kinetic energy is expressed as E = 0.5 * M * V^2. So the energy of the spacecraft when initially observed is 
E = 0.5 * M * V^2
 E = 0.5 * 2 kg * (4 m/s)^2
 E = 1 kg * 16 m^2/s^2
 E = 16 kg*m^2/s^2
 E = 16 J
 
 Later the energy is
 E = 0.5 * M * V^2
 E = 0.5 * 2 kg * (9 m/s)^2
 E = 1 kg * 81 m^2/s^2
 E = 81 kg*m^2/s^2
 E = 81 J
 
 So while falling towards the asteroid, it gained (81 J - 16 J = 65 J) of energy over a distance of 180 meters. Since work is defined as force times distance, the force exerted upon that spacecraft is
 65 J / 180 m
 = 65 kg*m^2/s^2 / 180 m
 = 0.361111111 kg*m/s^2
 = 0.361111111 N
 
 Dividing by the mass, gives
 = 0.361111111 N / 2 kg
 = 0.361111111 kg*m/s^2 / 2 kg
 = 0.180555556 m/s^2
 
 Now let's verify that answer. The equation for distance traveled with an initial velocity and under constant acceleration is
 d = VT + 0.5AT^2
 where
 d = distance
 V = initial velocity
 T = time
 A = acceleration
 
 Let's substitute the known values and see what we get for T
 d = VT + 0.5AT^2
 180 = 4*T + 0.5*0.180555556*T^2
 180 = 4*T + 0.090277778*T^2
 0 = 4*T + 0.090277778*T^2 - 180
 0.090277778*T^2 + 4*T - 180 = 0 
 Using the quadratic formula, I get T = -72, and T = 27.69230769.
 Since we're not dealing with time travel, we'll use the positive value of
27.69230769 seconds. 
 The velocity after 27.69230769 seconds should be the initial velocity plus the time multiplied by the acceleration, so
 4 m/s + 27.69230769 s * 0.180555556 m/s^2
 = 4 m/s + 5 m/s
 = 9 m/s 
 The velocity matches the acceleration, so the calculated acceleration of
0.180555556 m/s^2 is correct and when rounding to 2 significant figures is 0.18 m/s^2

If football was tossed in air without gravity or friction the ball would go into space