It is number 3 because I know it is
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
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Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
