a tool. The acceleration free-fall on the moon is 1/6 of its value on earth
Suppose he hit the ball with a speed of 27m/s at an angle 32 degree above the horizontal
How long was the ball in flight?
How far did it Travel?
How much farther would it travel on the moon than on earth?
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Posted in Golf Clubs by admin
Alright, the components of the ball’s initial velocity are, for the horizontal, 27m/scos32°, and 27m/ssin32° for the vertical. The final speed in the horizontal is the same as the initial because there is zero acceleration in that direction. Since we can assume that the ball lands vertically at the same height it was hit from, its displacement is zero as well. The time of flight is:
Δy = v₀t + 0.5gt²
Since we determined that vertical displacement is zero, this becomes:
0 = v₀t + 0.5gt²
gt² = 2v₀t
t = 2v₀ / g
= 2(27m/ssin32°) / (9.8m/s² / 6)
= 18s (rounded)
The distance it traveled may be found from the horizontal components of the ball’s velocity:
Δx = (v₀ + v)t / 2
= (27m/scos32° + 27m/scos32°)18s / 2
= 410m (rounded)
On earth, the time of flight will be different, it is:
t = 2v₀ / g
= 2(27m/ssin32°) / 9.8m/s²)
= 2.9s
So, by the same method as before, its range is:
Δx = (v₀ + v)t / 2
= (27m/scos32° + 27m/scos32°)2.9s / 2
= 66m (rounded)
Hope this helps.