On the apollo 14 mission to the moon, astronaut alan shepard hit a golf ball with a golf club improvised from a tool. the free-fall acceleration on the moon is 1/6 of its value on earth. suppose he hit the ball with a speed of 35 m/s at an angle 27 ∘ above the horizontal. part a how long was the ball in flight? express your answer using two significant figures.

On the moon, the gravitational acceleration is 1/6 of 9.8 m/s², so
g = 9.8/6 = 1.633 m/s²

Launch speed = 35 m/s
Launch angle = 27°  above the horizontal.
Therefore,
The horizontal velocity is
u = 35*cos(27) = 31.1852 m/s
The vertical launch velocity is
v = 35*sin(27) = 15.8897 m/s

Part A
When the ball reaches maximum height, the time requires is given by
0 = v - gt
t = v/g = 15.8897/1.6333 = 9.7286 s
This is one half of the time of flight, which is
2*9.7286 = 19.457 s

19.46 s (2 sig. figs)

stands for electromotive force and it is measured in volts. despite its name, it is not actually the force in the sense of being a "pushing" or "pulling" (not measured in newtons).an equation that relates emf,  ℰ,  to a battery's terminal voltage, v, current, i, and internal resistance, r, is    ℰ = v + irin this case, the emf can be interpreted as a voltage—a difference in electric potential.i would not say that it would be considered an energy. if emf can be defined as the energy/work per a unit charge from a battery or other electric source, then it is not strictly energy.