A particle is projected from the ground with a kinetic energy E at an angle of 60 degrees with the horizontal. Its kinetic energy at the highest point of its motion will be:
Correct answer: C. E/4
- A. E
- B. E/2
- C. E/4
- D. E/8
Explanation
At the highest point of its motion, the projectile will only have the horizontal component of its initial velocity. The vertical component will be zero. Therefore, the kinetic energy at this point will be equal to the kinetic energy associated with the horizontal component of its initial velocity.The horizontal component of the initial velocity can be found as:v₀x = v₀ cos θwhere v₀ is the initial velocity magnitude, and θ is the angle with the horizontal.Therefore, the kinetic energy at the highest point can be found as:E° = (1/2) m v₀x² = (1/2) m (v₀ cos θ)² = (1/2) m v₀² cos² θSince θ = 60°, we have:E° = (1/2) m v₀² (1/4) = (1/4) EThus, the kinetic energy at the highest point is E/4.
This question appeared on the NUMS 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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