How Does Kinetic Energy Change at Maximum Height in Projectile Motion?

[Brianne]

kinetic energy stuff...

here's the question...
an object is thrown upwards at a projection angle of 60 degree with kinetic energy E. When the object reaches max height, its kinetic energy becomes
A. (1/8)E
B. (1/4)E
C. (1/2)E
D. (3/2)E

I'm totally no idea...some1 kindly tells me the concept of this question... x_x

 

[joe tomei]

When the object is just launched it has E Cos (60) energy in the horizontal direction and E Sine (60) in the vertical direction. When it arrives at the top end of travel (vertically) it only has the horizontal energy left.
Therefore it is E Cos (60) = 0.5 E

 

[Doc Al]

You need to know what percentage of the object's initial KE is associated with the horizontal component of its velocity. (The part associated with the vertical component will go to zero at max height.) Hint: If the initial speed is V, what's the horizontal component of the velocity?

 

[Brianne]

i've seen the answer is 0.25E from the book, so I just think should we squaring the cos60, ie (cos60)^2=0.25?

Doc Al: if the horizontal component of velocity remains unchange in a projectile motion, so the velocity is still v isn't it?

 

[Dr.Brain]

Initial Kinetic energy : E = \frac{1}{2}mv^2

K.E at the highest position: \frac{1}{2}m(v cos60)^2

Rest i leave upto you.

BJ

 

[gnpatterson]

Brilliant question, makes you see _why_ KE _is_ proportional to velocity squared. If you think about KE has to be, then, this question reveals that it must be proportional to velocity squared simply because it is separable in the different spatial directions. WOW it really blows my mind. It is probably neater to say that the velocity in the horizontal direction is half the initial straight line velocity from simple geometry rather than having to appeal to trig (though trig is more general). And then say it this half velocity that you are squaring to get the quarter energy.

 

[Doc Al]

i've seen the answer is 0.25E from the book, so I just think should we squaring the cos60, ie (cos60)^2=0.25?

Yes, but why? The answer is to compare the initial KE (which comes from both the vertical and horizontal components of the initial velocity) to the final KE (which comes only from the horizontal speed). (See Dr.Brain's post.)

Doc Al: if the horizontal component of velocity remains unchange in a projectile motion, so the velocity is still v isn't it?

I'm not sure what you mean. If the initial velocity (with magnitude V) has components:
V_{ix} = V \cos \theta
V_{iy} = V \sin \theta

Then at the top of the motion, since the vertical component is zero, the velocity will have components:
V_{fx} = V \cos \theta
V_{fy} = 0