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

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The discussion revolves around calculating the kinetic and potential energy of a 2 kg ball dropped from a 40-meter cliff at an initial speed of 24 m/s at a 30-degree angle. The initial kinetic energy is calculated using the formula Ek1=(m*v^2)/2. At the highest point of its trajectory, the kinetic energy becomes zero, while potential energy needs to be calculated based on the height reached. The kinetic energy upon hitting the ground can be determined using conservation of energy principles, factoring in both vertical and horizontal components of velocity. The final calculations require careful consideration of height measurements and the application of equations of motion.
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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?
 
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Elaia06 said:
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.
 
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 ?
 
I'd calculate it from the ground at the bottom of the cliff (eg add 40m) because that's where it will end up.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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