A shot is fired at an angle of 60o to the horizontal with kinetic energy E. If air resistance is ignored, the kinetic energy at the top of the trajectory is:
Correct answer: C. E/4
- A. Zero
- B. E/8
- C. E/4
- D. E/2
Explanation
K.E=E=1/2mv2K.E=1/2mvx2 (horizontal velocity)K.E= 1/2m x (vcos 60)2K.E=1/2mv2 (0.5)2K.E=1/2mv2 (o.25)K.E=1/2mv2 (¼)K.E= ¼ ( ½ mv2)K.E = ¼ (E)K.E= E/4 ans
This question appeared on the ETEA MDCAT 2014 paper, which you can sit online with every answer explained.
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About Projectile Motion
Projectile motion combines uniform horizontal motion with vertically accelerated motion under gravity, assuming air resistance is neglected. Questions use the components of initial velocity to find time of flight, maximum height, horizontal range and position, while distinguishing projectile motion from general circular or one dimensional motion.
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