A golfer hits a shot to a green that is elevated 3.0 m above the point where the

Question

A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of 18.4 m/s at an angle of 55.0

Explanation / Answer

The intitial vertical velocity is 18.7sin55=18.7 * 0.819 =15.318 m/s. The maximum height of the projected motion is where vertical velocity =0 v^2=u^2 + 2as using a=g=>9.81 0=15.318^2 + 2 *(-9.81)*s Note the negative in parenthesis because if the motion is upwards, you are decelerating with gravity 2*9.81*s =234.646 s=234.646/(2*9.81) s=11.96m This is max height. However, it ends at 3m above. The difference in height =11.96-3=8.96 Using the same equation v^2=u^2 + 2as using a=g=>9.81 where u, the intial velocity=0 v^2=0+2*9.81*8.96 v^2=175.7592 v=v175.7592 v=13.25 m/s This is the velocity before it hits the green.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.