A body is projected vertically upwards with initial velocity u. The time taken to reach maximum height is:
A. u/g
B. g/u
C. u/(2g)
D. 2u/g
Answer: Option A
Solution (By JKSSB Mock Tests)
At maximum height, final velocity v = 0. Using v = u - gt (taking upward positive), 0 = u - gt ⇒ t = u/g. This is time of ascent. Total time of flight would be 2u/g. The equation derives from Newton's first equation of motion under constant acceleration g downward. Memory aid: Time to peak = initial velocity / gravitational acceleration. This fundamental result appears in projectile motion problems. Competitive exams often combine this with energy conservation or symmetry concepts for advanced questions.
Explanation:
Heat for phase change Q = m·L_f, where L_f is latent heat of fusion. Thus Q = 2 kg × 336 kJ/kg = 672 kJ. This direct application tests calorimetry fundamentals. Memory aid: 'Latent heat: Q = mL; no temperature change during phase transition'. Competitive exams frequently test such calculations with standard values. Always ensure units match (kg and kJ/kg here); convert if necessary. This problem assesses basic formula application skills essential for thermodynamics sections.
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