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777dan777 [17]
2 years ago
8

A player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

Physics
1 answer:
____ [38]2 years ago
7 0

At the highest point in its trajectory, the ball's acceleration is zero but its velocity is not zero.

<h3>What's the velocity of the ball at the highest point of the trajectory?</h3>
  • At the highest point, the ball doesn't go more high. So its vertical velocity is zero.
  • However, the ball moves horizontal, so its horizontal component of velocity is non - zero i.e. u×cosθ.
  • u= initial velocity, θ= angle of projection

<h3>What's the acceleration of the ball at the highest point of projectile?</h3>
  • During the whole projectile motion, the earth exerts the gravitational force with a acceleration of gravity along vertical direction.
  • But as there's no acceleration along vertical direction, so the acceleration along vertical direction is zero.

Thus, we can conclude that the acceleration is zero and velocity is non-zero at the highest point projectile motion.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

A- neither the ball's velocity nor its acceleration are zero.

B- the ball's acceleration points upward.

C- the ball's acceleration is zero but its velocity is not zero.

D- the ball's velocity points downward.

Learn more about the projectile motion here:

brainly.com/question/24216590

#SPJ1

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Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

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Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

tangent( O ) = \frac{opposite}{adjacent}

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

tangent( 4) = \frac{53 feets}{DB}

DB=  \frac{53 feets}{tangent( 4)}

DB=  757,93 feets

This its equal to the distance from the boat to the shoreline.

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