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

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7 0

2 years ago

John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2

Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function $cos(\omega t +\phi)$ is defined between -1 and 1, So it is not possible obtain a value $2\pi$ greater.

In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have $f=1+cos(\omega t +\phi)$ , the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the $f=1+cos(\omega t +\phi)$ , and its mean that the function $f=cos(\omega t +\phi)$ , in general, is not greater than $cos(\omega t +\phi)$ .

3 0

3 years ago