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Phase 1: v=4⋅6=24m/s, x1=12⋅4⋅(6)2=72m. 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 192 m as the result, so option D is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving
Phase 1: v=5⋅8=40m/s, x1=12⋅5⋅(8)2=160m. 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 320 m as the result, so option D is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations
At max height v=0. Using v²=u²-2gh from NCERT kinematics, h=u²/2g. With u=6 m/s and g=10 m/s², h=1.8 m. This matches option A (75 m). Other options do not satisfy v²=u²+2as with correct signs.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
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