Calculating Kinetic and Potential Energy of a Dropped Ball | Energy Question

[Questions999]

A ball which has a mass of 2 kg is dropped vertically with the initial speed of 24 m/s in the direction that forms the angle 30 degree with the horizon ,from a 40 meters tall cliff. Find a) The initial kinetic energy of the ball
My solution is Ek1=(m*v^2)/2 here i replace m=2 kg and v=24 m/s and i find Ek1.

b) its kinetic energy and potential energy when he is in the highest point of its trajectory.

When he is at the highest point, v=0 so the Kinetic energy is zero..as for the potential energy..what am I supposed to do?

c) Find the kinetic energy of the ball when he touches the ground... the kinetic energy supposed to be E=(m*v^2)/2 ,by the law of the conservation of the energy, I have that Ep(b)=Ekc+Epc where Epb is the potential energy in the point B,and Ekc and Epb are the Kinetic and potential energy in the point c
d)Find the speed of the ball when it hits the ground...Here Ep=0..I have to find Ek but how?

 

[CWatters]

A ball which has a mass of 2 kg is dropped vertically with the initial speed of 24 m/s in the direction that forms the angle 30 degree with the horizon ,from a 40 meters tall cliff.

"dropped vertically" ?? Did you really mean fired up at an angle of 30 degrees to the horizontal?

If so then..

b) Work out the vertical component of it's initial velocity. Apply standard equations of motion to work out how high it goes. Calculate the PE but think carefully about where the hieght needs to be measured from. As for it's KE.. At the top the vertical component is indeed zero but it will still have same horizontal component so it still has some KE.

c) Two ways to approach this: 1) You could apply the same equations of motion and answer part d first. Then calculate the KE. or 2) Use the answers from b) and apply conservation of energy.

d) It falls from the height calculated in b). Apply standard equation of motion to calculate the vertical velocity when it hits the ground. Horizontal velocity still the same as that at launch. Vector addition.

 

[Questions999]

For the point b,i won't measure it from the ground right? I have to measure it from the point it is being thrown ?

 

[CWatters]

I'd calculate it from the ground at the bottom of the cliff (eg add 40m) because that's where it will end up.

 

[Nickie]

a) The initial kinetic energy of the ball can be calculated using the formula Ek = (1/2) * m * v^2, where m is the mass of the ball and v is the initial velocity. Substituting the values given, we get Ek = (1/2) * 2 kg * (24 m/s)^2 = 576 J.

b) At the highest point of its trajectory, the ball has zero kinetic energy since it is momentarily at rest. Its potential energy can be calculated using the formula Ep = m * g * h, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ball from the ground. Substituting the values given, we get Ep = 2 kg * 9.8 m/s^2 * 40 m * cos(30 degrees) = 392 J.

c) When the ball touches the ground, its potential energy is zero since it is at the same height as the ground. Using the law of conservation of energy, we can equate the initial kinetic energy (Ek1) to the final kinetic energy (Ekc) and potential energy (Epc) at the ground. This gives us the equation Ek1 = Ekc + Epc. Substituting the values from part a and b, we get Ek1 = (1/2) * 2 kg * v^2 = 576 J, Ekc = (1/2) * 2 kg * v^2 = 576 J, and Epc = 0. Solving for v, we get v = √576 * 2 / 2 kg = 24 m/s.

d) To find the speed of the ball when it hits the ground, we can use the same equation from part c and solve for v. This gives us v = √(2 * 576 J) / 2 kg = √576 = 24 m/s. Therefore, the speed of the ball when it hits the ground is 24 m/s.