Predicted height and total energy

The predicted height is 2.809 meters, writing this in centimeters we get (1m = 100cm):

h = 2.809 m = (2.809)*(100cm) = 280.9 cm

And the total energy is:

E = 6.696 J

Explanation:

First let's see the problem.

We have an object of mass m = 274g which is thrown upwards with an initial velocity v0 = 6.991 m/s, in a place with a gravitational acceleration of g = 8.7 m/s^2

When the object is on the air, the only force acting on it will be the gravitational force, then the acceleration of the object will be equal to the gravitational acceleration, then we can write:

a(t) = -8.7 m/s^2

Where the negative sign is because this acceleration points down.

Now to get the velocity of the object we can integrate over time to get:

v(t) = (-8.7 m/s^2)*t + v0

Where v0 is a constant of integration, which is the initial velocity, then we can write this as:

v(t) = (-8.7 m/s^2)*t + 6.991 m/s

Now we can integrate again over the time to get the position equation.

p(t) = (1/2)*(-8.7 m/s^2)*t^2 + (6.991 m/s)*t + p0

Where p0 is the initial position, because the ball is being thrown from the ground, the initial position is 0.

Then the position equation is:

p(t) = (1/2)*(-8.7 m/s^2)*t^2 + (6.991 m/s)*t

Ok, now we know all the movement equations for the object.

The first thing we want to know is the maximum height of the object.

We know that the object reaches its maximum height when the velocity is zero (this is, the velocity stops being positive, meaning that the object stops going up, then in that time we have the maximum height)

We need to solve:

v(t) = 0m/s = (-8.7 m/s^2)*t + 6.991 m/s

(8.7 m/s^2)*t =  6.991 m/s

t =  6.991 m/s/( (8.7 m/s^2)  = 0.804 seconds

The maximum height of the object is given by:

p(0.804s) = (1/2)*(-8.7 m/s^2)*(0.804)^2 + (6.991 m/s)*(0.804) = 2.809 m

The maximum height of the object is 2.809 meters.

Now let's find the maximum energy.

Remember that the energy of an object can be written as the sum of the potential energy U and the kinetic energy K.

E = K + U

Such that for an object of mass m and velocity v, the kinetic energy is:

K = (1/2)*m*v^2

And for an object of mass m, at a height h from the ground and with gravitational acceleration g, the potential energy is:

U = m*g*h

Now, when the object is at its maximum height, the velocity is zero.

Then K = 0

And for conservation of energy, the total energy of the object becomes potential energy.

E = 0 + U

E = U

So if we find the potential energy at the maximum height of the object's path, we can find the total energy of the object.

We know that:

mass = m = 274g = 0.274 kg  (here i used that 1kg = 1000g)

height = h = 2.809 meters.

gravitational acceleration = g = 8.7 m/s^2

Then the potential energy at this point is:

U =  0.274 kg*(2.809 meters)*(8.7 m/s^2) = 6.696 J

This means that the total energy of the object is:

E = 6.696 J

c. magnitude and direction