A ball is thrown at an angle of 30 degrees to the horizontal. The initial kinetic energy of the ball is K. Air resistance has negligible effect on the motion of the ball. What is the kinetic energy of the ball at the maximum height?

A) 0

B) 0.25 K

C) 0.75 K

D) 0.85 K

___________________________________________

The answer is C according to the mark scheme.

I thought it was A because:

K=E=mv^2/2.

The m and 2 are constant.

So, K is proportional to velocity squared.

The vertical component of the velocity at the maximum would be 0 so K would be 0.

But I was royally incorrect and the answer was C.

So I thought maybe one has to work out the horizontal velocity at the max height.

I have no clue how to do that.

If anyone knows how to work this out I would appreciate your help.

Thanks.

A) 0

B) 0.25 K

C) 0.75 K

D) 0.85 K

___________________________________________

The answer is C according to the mark scheme.

I thought it was A because:

K=E=mv^2/2.

The m and 2 are constant.

So, K is proportional to velocity squared.

The vertical component of the velocity at the maximum would be 0 so K would be 0.

But I was royally incorrect and the answer was C.

So I thought maybe one has to work out the horizontal velocity at the max height.

I have no clue how to do that.

If anyone knows how to work this out I would appreciate your help.

Thanks.

(edited 6 years ago)

Consider the components involved and resolve them. At maximum height, the vertical component will be a specific value, as constrained by you, (0), and hence the resolution allows you to consider the horizontal component and solve

That being said, I end up getting C... are you sure it wasn't 30 degrees to the vertical? Or am I just failing at answering? ;-;

Either way, the diagram should help with resolving it

Hey, thanks for the help. The answer was indeed C, I misread it so apologises. I tried to resolve it but I'm not doing it right since I'm not getting the answer. I'd appreciate it if you could show how you solved. By the way, the angle was to the horizontal not to the vertical.

(edited 6 years ago)

I still don't know how they got the answer to be C. Any worked solution would be much appreciated!!

Original post by asifmahmoud

I still don't know how they got the answer to be C. Any worked solution would be much appreciated!!

Initial kinetic energy = 1/2 mv^2

At maximum height, vertical component is zero. So speed is only the horizontal component. So kinetic energy = 1/2 m(vcos30)^2 = 3/8 mv^2

Ratio = 3/8 / 1/2 = 3/4

Initial kinetic energy = 1/2 mv^2 = K

3/8 mv^2 = 3/4 x 1/2mv^2 = 3/4 K so answer 0.75K C

Original post by jmushtaq

Hey, thanks for the help. The answer was indeed C, I misread it so apologises. I tried to resolve it but I'm not doing it right since I'm not getting the answer. I'd appreciate it if you could show how you solved. By the way, the angle was to the horizontal not to the vertical.

5 Years late but this is for a new student I guess lol. It's total KE is 1.0K At the beginning that is split between horizontal and vertical. But at max height there is no vertical component and no air resistance means horizontal velocity is constant. So at max height it's just cos(30) squared K. Which is 3/4K.

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