Questions and explanations focused on free fall and vertical motion under gravity. Covers problems involving objects moving upward or downward, including time of flight and velocity.
Average speed = total distance / total time. 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 3 m/s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
Time to max height: v=v0+at, 0=48−10t⇒t=4.8s. 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 4.8 s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
Phase 1: v2=v02+2ax, 252=102+2a(150)⇒625=100+300a⇒a=1.75m/s2. 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 s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
Use v2=v02+2ax. Here, v=0, v0=18m/s, x=36m. 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 36 m/s² as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
Speed: 72km/h=20m/s. 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 20 m/s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
For uniformly accelerated motion, distance is given by x=v0t+12at2. 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 100 m as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
Phase 1: 8=20−4t⇒t1=3s. 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 6 s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
Use v2=v02+2ax. Here, v0=0, v=24m/s, x=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 3 m/s² as the result, so option C is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
Find a=v−v0t=12−04=3m/s2. 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 24 m as the result, so option B is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
At max height v=0. Using v²=u²-2gh from NCERT kinematics, h=u²/2g. With u=15 m/s and g=10 m/s², h=11.2 m. This matches option A (11.2 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
Displacement y=−30m, −30=30t−5t2⇒5t2−30t−30=0⇒t2−6t−6=0. 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 6.87 s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
Use x=v0t+12at2. 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 54 m as the result, so option B is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
