Ball thrown up with an initial speed of 27.0 m/s

In summary: I'm not sure why.In summary, the ball's potential energy is constant, the velocity and acceleration vectors are in the same direction at t=7.3s, acceleration is constant, acceleration is not 0 near t=7.3s, maximum height is not reached at t=7.3s, and the ball's kinetic energy increases during the first second.
  • #1
evan b
10
0

Homework Statement


A) The ball's PE is a constant.
B) The velocity vector and the acceleration vector are in the same direction after t= 7.3s.
C) The acceleration is constant.
D) The acceleration is 0 near t= 7.3s.
E) Maximum height is reached at about t=7.3s.
F) The ball's KE increases during the first second.

The Attempt at a Solution


I am pretty sure of the following. Somewhere i am wrong.
A is false
B is true (given once it hits 7.3, gravity is negative(acceleration) and the direction is necative
C is true because accel is gravity and that is always constant.
D is false because if the gravity is constant than it wouldn't be zero
E is true because it starts coming down after 7.3 s
F is true because KE is zero when the ball is in hand but when he releases it, it gains
 

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  • #2
I can't see the image until it's approved, but this is my input

A - I would agree that this is false.
B - mathematically I would agree, however I would say that the ball has long since hit the ground at this point. (possibly numbers are wrong)
C - I would agree that this is true.
D - I agree that this is false.
E - I would disagree here, how did you come to this conclusion? It looks more like 2.75 seconds to me. Unless you've stated some of the numbers wrong/I've misinterpreted them.
F - kinetic energy of course being: .5mv^2 Think about which way the ball is accelerating, think of if v is increasing or decreasing.

EDIT: latex isn't working for me
 
  • #3
KE and PE decreases.

I can confirm that your responses are mostly correct. A is false because the ball's potential energy is constantly changing as it moves through the air. B is true because the velocity and acceleration vectors are both pointing downwards after t=7.3s. C is also true because the acceleration due to gravity is a constant force acting on the ball. D is false because the acceleration is not zero at any point during the ball's motion. E is true because the ball reaches its maximum height at t=7.3s before starting to fall back down. F is also true because the ball's kinetic energy increases as it gains speed during its first second of motion. Overall, your understanding of the concepts is correct, but be careful with your wording and make sure to use proper units when discussing quantities like velocity and acceleration.
 

What is the initial velocity of the ball?

The initial velocity of the ball is 27.0 m/s. This means that when the ball is first thrown, it is moving at a speed of 27.0 meters per second in the upward direction.

How high does the ball go?

The maximum height that the ball will reach can be calculated using the equation h = (v02sin2θ)/2g, where v0 is the initial velocity, θ is the angle at which the ball is thrown, and g is the acceleration due to gravity (9.8 m/s2).

How long does it take the ball to reach its maximum height?

The time it takes for the ball to reach its maximum height can be calculated using the equation t = v0sinθ/g, where v0 is the initial velocity and θ is the angle at which the ball is thrown.

What is the velocity of the ball at its maximum height?

The velocity of the ball at its maximum height is 0 m/s. This is because at the highest point of the ball's trajectory, it has reached its maximum height and is no longer moving vertically.

What happens to the ball after it reaches its maximum height?

After reaching its maximum height, the ball will start to fall back down towards the ground due to the force of gravity. Its velocity will increase until it reaches the ground, assuming there are no other forces acting on it.

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