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.
Answers
Answered by
3
Final Answer :
1 : 1.763m
2: 70.31° North of East.
Steps and Understanding :
1) Convention :
As shown in diagram
Upward and rightward direction vector is positive.
Downward direction and leftward vector is negative.
2) Let A be the initial position of golfer.
From here, golfer went on his first put towards B, then C and then D.
3) We have to find AD vector .
4) We will find first all vectors, AB,BC,CD to get AD with appropriate sign.
5) Then, from AD we will get it's magnitude and direction .
1 : 1.763m
2: 70.31° North of East.
Steps and Understanding :
1) Convention :
As shown in diagram
Upward and rightward direction vector is positive.
Downward direction and leftward vector is negative.
2) Let A be the initial position of golfer.
From here, golfer went on his first put towards B, then C and then D.
3) We have to find AD vector .
4) We will find first all vectors, AB,BC,CD to get AD with appropriate sign.
5) Then, from AD we will get it's magnitude and direction .
Attachments:
Answered by
3
Given Conditions ⇒
Golfer displaces the ball 3.66 m in north direction.
∴ Vector in this case (a₁) = 3.66 j.
Now, Golfer displaces the ball 1.83 m in South East direction.
∴ Vector in this case (a₂) = 1.83 Cos 45° - 1.83 Sin 45° j.
Golfer again displaces the ball 0.99 m in the South west direction.
∴ In Vector form (a₃) = -0.99 Cos 45° - 0.99 Sin 45° j.
Refers to the attachment for the better understanding.
I) Now,
∵ 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
∴ Magnitude of the Displacement =
=
=
=
= 1.765 m.
Hence, 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.
Attachments:
simonhamudombe:
thank you so much
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