Question

x direction. A golfer hits his tee shot a distance of 300.0 m, corresponding to a displacement of r1 = 300.0 mi and then hits his second shot 189.0 m corresponding to a displacement of r2 = 172.0 mi + 80.3 mj. What is the final displacement of the golf ball from the tee? Give your answer as a resultant vector that incorporates horizontal and vertical components.The 18th hole at Pebble Beach Golf Course is a dogleg to the left of length 496.0 m. The fairway off the tee is taken to be the x direction. A golfer hits his tee shot a distance of 300.0 m, corresponding to a displacement Δr⃗ 1=300.0miˆ,and hits his second shot 189.0 m with a displacement Δr⃗ 2=172.0miˆ+80.3mjˆ.What is the final displacement of the golf ball from the tee?

Answer 1

Answer:

(472i + 80.3j) m

Explanation:

Given the following :

Distance of tee shot = 300m

Distance of second shot = 189.0 m

Displacement r1 of tee shot :

r1 = 300 mi

Displacement r2 of second shot :

r2 = 172.0 mi + 80.3 mj

The final displacement of the golf ball from the tee:

r_final = r1 + r2

r_final = (300i)m + (172.0i + 80.3j) m

r_final = (300 + 172)i m + 80.3j m

r_final = (472i + 80.3j) m