A body is projected vertically upwards with initial velocity u. The time taken to reach maximum height is: MCQ with Answer and Explanation
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Question #1
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Two bodies A and B start from the same point with velocities 10 m/s and 15 m/s respectively in the same direction. If B starts 2 seconds after A, the time after which B catches A is:
Correct Answer: Option C
Explanation:
When B starts, A has already traveled 10 m/s × 2 s = 20 m. Relative velocity of B w.r.t A = 15 - 10 = 5 m/s. Time to cover 20 m gap at 5 m/s relative speed = 20/5 = 4 s after B starts. Since B started 2 s late, total time from A's start = 2 + 4 = 6 s. Alternatively, equate distances: 10t = 15(t-2) ⇒ 10t = 15t - 30 ⇒ t = 6 s. This relative motion approach simplifies pursuit problems. Memory tip: In same-direction motion, use relative velocity = |v₁ - v₂|. Frequently tested in kinematics sections of competitive exams.
When B starts, A has already traveled 10 m/s × 2 s = 20 m. Relative velocity of B w.r.t A = 15 - 10 = 5 m/s. Time to cover 20 m gap at 5 m/s relative speed = 20/5 = 4 s after B starts. Since B started 2 s late, total time from A's start = 2 + 4 = 6 s. Alternatively, equate distances: 10t = 15(t-2) ⇒ 10t = 15t - 30 ⇒ t = 6 s. This relative motion approach simplifies pursuit problems. Memory tip: In same-direction motion, use relative velocity = |v₁ - v₂|. Frequently tested in kinematics sections of competitive exams.
Question #2
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Projectile horizontal range maximum at angle:
Correct Answer: Option B
Explanation:
Range R = u²sin(2θ)/g. sin(2θ) maximum (=1) when 2θ=90° ⇒ θ=45°. Standard result for level ground projection. Memory aid: 'Max range at 45°; max height at 90°'. Projectile motion fact frequently tested in competitive exams.
Range R = u²sin(2θ)/g. sin(2θ) maximum (=1) when 2θ=90° ⇒ θ=45°. Standard result for level ground projection. Memory aid: 'Max range at 45°; max height at 90°'. Projectile motion fact frequently tested in competitive exams.
Question #3
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The energy stored in an inductor of inductance L carrying current I is
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