Finding ymax and ymin of a ball thrown straight into the air

[babygiggles85]

Homework Statement

A 300g ball is thrown straight up into the air. It is exactly 2m above the ground when it is released. It reaches a height of 12m above the height from which is was released, and then falls straight back down.

a) assume that y=0 is located at the point of release of the ball. Find the maximum and minimum values of y and PE(g). What is the total energy of the system? Find the maximum and minimum values of KE.

b) repeat part (a) with y=0 located on the ground

Homework Equations

I'm having trouble finding what equations to use. I tried using mgh + 1/2m(vf^2-vi^2) but the problem doesn't give velocity

The Attempt at a Solution

 

[Raziel2701]

Here's the thing about velocity, when the ball reaches its maximum altitude, its velocity is zero. You might need to employ kinematic equations for this one. Also mind the direction of gravity. When the ball is going up, gravity is pulling down.

Regarding the energy of the system, this is a conservation of energy problem. When the ball is at the top of its trajectory, when v equals zero, the ball has the highest amount of potential energy and no kinetic energy.

So use the fact that $$V_{final}=0$$ to solve for the highest point it reaches and always keep in mind the direction of the acceleration (g). Potential is highest at top of trajectory, KE is zero there too.

 

[tomcue]

I would first clarify any uncertainties with the given information. In this case, it is important to confirm whether the ball is being thrown from the ground or from a height of 2m. Assuming that the ball is thrown from a height of 2m, we can use the following equations to find the maximum and minimum values of y (position) and PE(g) (potential energy due to gravity):

y_max = 12m (given)
y_min = 2m (given)
PE(g)_max = mgy_max = (0.3kg)(9.8m/s^2)(12m) = 35.28 J
PE(g)_min = mgy_min = (0.3kg)(9.8m/s^2)(2m) = 5.88 J

To find the total energy of the system, we can use the conservation of energy principle, which states that the total energy (E) of a system remains constant:

E = PE(g) + KE

Since we know the values of PE(g)_max and PE(g)_min, we can find the total energy at those points:

E_max = PE(g)_max = 35.28 J
E_min = PE(g)_min = 5.88 J

To find the maximum and minimum values of KE (kinetic energy), we can use the formula:

KE = 1/2mv^2

However, as mentioned in the problem, the velocity (v) is not given. Without the velocity, we cannot accurately calculate the values of KE. Therefore, we would need to make assumptions or obtain more information to solve for KE.

In part (b) of the problem, if y=0 is located on the ground, the equations and calculations would be slightly different:

y_max = 10m (since the ball reaches a height of 12m above the ground, but y=0 is now located on the ground)
y_min = 0m (since the ball is released from the ground)
PE(g)_max = mgy_max = (0.3kg)(9.8m/s^2)(10m) = 29.4 J
PE(g)_min = mgy_min = (0.3kg)(9.8m/s^2)(0m) = 0 J

Again, without the information about velocity, we cannot accurately calculate the values of KE.