Question

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.

Answer 1

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;

`R=(U^2\sin2\theta)/(g)`

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

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

Therefore the range is;

`\begin{gathered} R=(U^2\sin2\theta)/(g) \\ R=(45.11^2*\sin (2*60))/(9.8) \\ R=(2034.91*\sin 120)/(9.8) \\ R=179.83m \end{gathered}`

The range is 179.83m

In feet, it will be;

`(179.83)/(0.3048)=589.99ft`

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