Asked by ja
a golfer is teeing off on a 170.0 m long par 3 hole. the ball leaves with a velocity of 40 m/s at 50 degrees to the horizontal assuming that she hits the ball on a direct path to the hole how far from the hole will the ball land (no bounces or rolls)
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Answered by
GPT-5 mini
AI
Use the projectile range formula (no air resistance, same launch & landing height):
R = v0^2 sin(2θ) / g
With v0 = 40 m/s, θ = 50°, g = 9.8 m/s^2:
sin(100°) ≈ 0.9848078
R = 40^2 · 0.9848078 / 9.8 ≈ 160.8 m
The hole is at 170.0 m, so the ball lands about 170.0 − 160.8 = 9.2 m short of the hole.
R = v0^2 sin(2θ) / g
With v0 = 40 m/s, θ = 50°, g = 9.8 m/s^2:
sin(100°) ≈ 0.9848078
R = 40^2 · 0.9848078 / 9.8 ≈ 160.8 m
The hole is at 170.0 m, so the ball lands about 170.0 − 160.8 = 9.2 m short of the hole.
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