Question

1. A golfer takes three putts to get the ball into the hole. The first put displaces the ball 3.66m north, the second 1.83m southeast, and the third 0.99m southwest. What are
I. Magnitude, and
II. The direction of the displacement needed to get the ball into the hole on the first putt.

Answer 1

Given Quantities ⇒

Firstly, Golfer displaces the ball by 3.66 m in North(N) direction.
∴ In Vector Form (a₁) = 3.66 j.

Secondly, Golfer displaces the ball by 1.83 m in South East (SE) direction.
∴  In Vector Form (a₂) = 1.83 Cos 45° - 1.83 Sin 45° j.
 
lastly, Golfer  displaces the ball by 0.99 m in the South West (SW) direction.
∴ In Vector form (a₃) = -0.99 Cos 45° - 0.99 Sin 45° j.

Look towards the attachment.  

I) According to the Question, we need to find the Magnitude of the Displacement.

∴ Total Displacement (a) = a₁ + a₂ + a₃
 = 3.66 j + 1.83 Cos 45°i - 1.83 Sin 45° j - 0.99 Cos 45° i - 0.99 Sin 45° j.
 = 3.66 j + 1.83 Cos 45° i - 0.99 Cos 45° i - 1.83 Sin 45° j - 0.99 Sin 45° j.
 = 3.66 j + Cos 45°i (1.83 - 0.99) - Sin 45°j (1.83 + 0.99)
 = 3.66 j + Cos 45°i (0.84) - Sin 45° j(2.82)
( ∵ Sin 45° = Cos 45° = $\frac{1}{ \sqrt{2} }$ )

 = 3.66 j + 1/√2 × 0.84 i - 1/√2 × 2.82 j
 =  1.66 j + 0.6 i

Thus, , Magnitude of the Displacement =  $\sqrt{1.66 j + 0.6 i}$
  a = $\sqrt{(1.66)^{2} + (0.6)^{2} }$
  a = √(2.7556 + 0.36)
  a = √3.1156
  a = 1.765 m.
∴ a ≈ 1.76 m.


Thus, the magnitude of the Displacement covered by the ball is 1.765 (≈ 1.76) m.

II) Let us find the Direction of the Displacement,
We know,
∴ tanθ = 1.66/0.6
∴ tanθ = 2.767
∴ tanθ = tan 70.31°
On Comparing, 
 θ = 70.31°

∴ Direction is at the angle of 70.31° (Approximately) in east.


Hope it helps.