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 ⇒

Golfer displaces the ball 3.66 m in N direction.
∴ Vector in this case (a₁) = 3.66 j.

Thus, Golfer displaces the ball 1.83 m in SE direction.
∴ Vector in this case (a₂) = 1.83 Cos 45° - 1.83 Sin 45° j.
 
At last, Golfer  displaces the ball 0.99 m in the SW direction.
∴ In Vector form (a₃) = -0.99 Cos 45° - 0.99 Sin 45° j.

See the diagram attached below. 

I) Now, therefore 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)
( ∵ Cos 45° = Sin 45° = 1/√2)

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

Therefore, Magnitude of the Displacement = $\sqrt{1.66j + 0.6 i}$
 ⇒ a = $\sqrt{(1.66)^{2} + (0.6)^{2} }$
 ⇒ a = $\sqrt{2.7556 + 0.36}$
 ⇒ a = $\sqrt{3.1156}$
 ⇒ a = 1.765 m.


Thus, the magnitude of the Displacement covered by the ball is 1.756 m.

II) Now, We need to find the direction of the Displacement,
∴ 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° (Approx) in east.


Hope it helps.