A golf ball is being struck by a club. The ball leaves the club with a speed of 148 feet per second at an angle of 60° with the horizontal. If the ball strikes the ground 6 seconds later, how far from the golfer does that ball land? Assume level ground and neglect air resistance. Round your answer to the nearest tenth.

A golf ball is being struck by a club. The ball leaves the club with a speed of 148 feet per second at an angle of 60° with the horizontal. If the ball strikes class=

Answer :

The motion of the golf ball is a projectile motion.

The horizontal distance the ball travels before striking the ground again is the range.

The formula to calculate range is;

[tex]R=\frac{U^2\sin2\theta}{g}[/tex]

where U = initial velocity = 148 ft/s

1 foot = 0.3048m

148 ft/s = 148 x 0.3048 = 45.11

Hence, U = 45.11 m/s

[tex]\begin{gathered} U=45.11\text{ m/s} \\ \theta=60^0 \\ g=9.8m/s^2 \end{gathered}[/tex]

Therefore the range is;

[tex]\begin{gathered} R=\frac{U^2\sin2\theta}{g} \\ R=\frac{45.11^2\times\sin (2\times60)}{9.8} \\ R=\frac{2034.91\times\sin 120}{9.8} \\ R=179.83m \end{gathered}[/tex]

The range is 179.83m

In feet, it will be;

[tex]\frac{179.83}{0.3048}=589.99ft[/tex]

Therefore, to the nearest tenth, the ball lands 590.0 feet or 179.8m from the golfer after being struck.

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