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
