How Do You Calculate Ymax and Velocity at Half Ymax Using Energy Conservation?

[student 1]

Homework Statement

A ball is thrown vertically from y=0 with velocity V₀. Using only energy considerations, find the following
a. Y_max in terms of V₀.
B. The Velocity(V_1/2 at y= 1/2 Y_max (in terms of V₀

Homework Equations

K_i+U_i=K_f+U_f

The Attempt at a Solution

I'm not sure how to start this off to find the y_max in terms of V₀

 

[Dick]

The U in your problem is the potential energy mgy. The K is the kinetic (1/2)mv^2. At the initial time v=V0 and y=0. At the final time v=0 and y=ymax. Use your energy conservation to solve for ymax in terms of V0.

 

[Moetasim]

. However, using energy considerations, we can determine the velocity at any point along the trajectory of the ball. At the initial point, the ball has only kinetic energy, as it has not yet gained any potential energy. As it travels upwards, it gains potential energy due to its position in the Earth's gravitational field. At the highest point of the trajectory, the ball has reached its maximum potential energy and has no kinetic energy. This means that at this point, its velocity is 0. Using the conservation of energy equation, we can set the initial kinetic energy equal to the final potential energy:

Ki = Kf + Uf
(1/2)mV0^2 = 0 + mgy
where m is the mass of the ball, g is the acceleration due to gravity, and y is the maximum height reached by the ball.

Solving for y, we get:
y = V0^2/2g

This is the maximum height reached by the ball in terms of its initial velocity V0.

To find the velocity at y = 1/2 y_max, we can use the same equation, but this time we are solving for V1/2:
(1/2)mV0^2 = (1/2)mv^2 + mg(1/2 y_max)
where v is the velocity at y = 1/2 y_max.

Solving for v, we get:
v = √(V0^2 - gy_max)

This is the velocity at y = 1/2 y_max in terms of V0.