A golfer takes two strokes to putt a golf ball into a hole. On the ﬁrst stroke, the ball moves 4.0m at an angle 60◦ East of North. On the second, it moves 3.0m at an angle 70◦ South of East. If the golfer had instead hit the ball directly into the hole on the ﬁrst stroke, what would have been the magnitude and direction of the ball’s displacement

Question

A golfer takes two strokes to putt a golf ball into a hole. On the ﬁrst stroke, the ball moves 4.0m at an angle 60◦ East of North. On the second, it moves 3.0m at an angle 70◦ South of East. If the golfer had instead hit the ball directly into the hole on the ﬁrst stroke, what would have been the magnitude and direction of the ball’s displacement

Answer 1

sketch it. labale the angles, You have a SAS triangle. You want the third side:

law of cosines
c^2=a^2+b^2 -2ab*cosC
you know, a, b, and angle C

Answer 2

All angles are measured CCW from +x-axis.

Disp. = 4m[30o] + 3[290].
X = 4*Cos30 + 3*Cos290 = 4.49 km.
Y = 4*sin30 + 3*sin290 = -0.82km.

x + yi = 4.49 - 0.82i = 4.56km[-10.3] =
4.56km[10.3o]S. of E.

Answer 3

To solve this problem, we will use vector addition to find the resultant displacement if the golfer had hit the ball directly into the hole on the first stroke.

First, let's break down the given information into its vector components.

For the first stroke:
Magnitude: 4.0m
Angle: 60° East of North

To find the horizontal and vertical components, we can use trigonometry:

Horizontal Component: 4.0m * sin(60°) = 3.46m
Vertical Component: 4.0m * cos(60°) = 2.0m

So, the first stroke has a horizontal component of 3.46m and a vertical component of 2.0m.

For the second stroke:
Magnitude: 3.0m
Angle: 70° South of East

To find the horizontal and vertical components, we can again use trigonometry:

Horizontal Component: 3.0m * cos(70°) = 0.904m
Vertical Component: 3.0m * sin(70°) = 2.78m (Note: Since it is South of East, the vertical component is positive.)

So, the second stroke has a horizontal component of 0.904m and a vertical component of 2.78m.

Now, let's add the components of the two strokes together to find the total horizontal and vertical displacement:

Horizontal Displacement = 3.46m + 0.904m = 4.364m
Vertical Displacement = 2.0m + 2.78m = 4.78m

We can use the Pythagorean theorem to find the magnitude of the resultant displacement:

Magnitude = √(Horizontal Displacement^2 + Vertical Displacement^2)
= √(4.364m^2 + 4.78m^2)
= √(19.036m^2)
= 4.36m

Therefore, if the golfer had hit the ball directly into the hole on the first stroke, the magnitude of the ball's displacement would have been approximately 4.36m.

To find the direction of the ball's displacement, we can use trigonometry again:

Direction = tan^(-1)(Vertical Displacement / Horizontal Displacement)
= tan^(-1)(4.78m / 4.364m)
= 49.1°

Therefore, the direction of the ball's displacement would have been approximately 49.1°.