Determine the resultant moment produced by the load and the weights of the tower crane jibs about point A and about point B. Exp
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Answer:
The ball is in the air for 3.5 seconds
The initial horizontal component of velocity is 28.6 m/s
The vertical component of the final velocity is 34.3 m/s downward
The final velocity is 44.7 m/s in the direction 50.2° below the horizontal
Explanation:
A ball is thrown horizontally
That means the vertical component of the initial velocity
The initial velocity is the horizontal component
The ball is thrown from the top of a 60 m
That means the vertical displacement component y = 60 m
→ y = t +
gt²
where g is the acceleration of gravity and t is the time
y = -60 m , g = -9.8 m/s² ,
Substitute these values in the rule
→ -60 = 0 + (-9.8)t²
→ -60 = -4.9t²
Divide both sides by -4.9
→ 12.2449 = t²
Take √ for both sides
∴ t = 3.5 seconds
* <em>The ball is in the air for 3.5 seconds </em>
The initial velocity is the horizontal component
The ball lands 100 meter from the base of the building
That means the horizontal displacement x = 100 m
→ x = t
→ t = 3.5 s , x = 100 m
Substitute these values in the rule
→ 100 = (3.5)
Divide both sides by 3.5
→ = 28.57 m/s
<em>The initial horizontal component of velocity is 28.6 m/s</em>
The vertical component of the final velocity is
→ =
+ gt
→ = 0 , g = -9.8 m/s² , t = 3.5 s
Substitute these values in the rule
→ = 0 + (-9.8)(3.5)
→ = -34.3 m/s
<em>The vertical component of the final velocity is 34.3 m/s downward</em>
The final velocity v is the resultant vector of and
→ Its magnetude is
→ Its direction
→ = 28.6 ,
= -34.3
Substitute this values in the rules above
→
→ Its direction
The negative sign means the direction is below the horizontal
<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>