A stone is thrown with an initial speed of 11.5 m/s at an angle of 50.0 above the horizontal from the top of a 30.0-m-tall build
1 answer:
Answer:
The magnitude of the horizontal displacement of the rock is 7.39 m/s.
Explanation:
Given that,
Initial speed = 11.5 m/s
Angle = 50.0
Height = 30.0 m
We need to calculate the horizontal displacement of the rock
Using formula of horizontal component
Put the value into the formula
Hence, The magnitude of the horizontal displacement of the rock is 7.39 m/s.
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Part a)
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Part b)
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Answer:
The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>
Explanation:
The directions in which the delivery truck travels are;
1) 2.8 km North = 2.8·, in vector form
2) 1.0 km East = 1.0·, in vector form
3) 1.6 km South = -1.6·, in vector form
Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;
Final displacement, Δd = 2.8· + 1.0·
+(-1.6·
) = 1.2·
+ 1.0·
Final displacement, = 1.0· + 1.2·
Where;
Δx = The displacement in the x-direction = 1.0·
Δy = The displacement in the y-direction = 1.2·
The magnitude of the resultant displacement vector is given as follows
= √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)
The magnitude of the resultant displacement vector ≈ 1.6 km
The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE
Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.