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Vertical motion: y=12gt2, where y=−39.2m. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 12 gt as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8
For free fall, s=½gt² and v²=2gh per NCERT. With g=10 m/s², given height or velocity yields time or distance via these equations. Substituting values gives result matching 20 m. This confirms option C as correct by uniformly accelerated motion equations.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8
Vertical motion: y=12gt2, where y=−31.25m. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 12 gt as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8
For vertical motion: y = v₀y t + (1/2) a\_y t², where v₀y = 0, a\_y = -g. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 2 s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving
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