11 public questions tagged with this topic.
Vertical motion: y=12gt2, where y=−45m. 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 s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion Parameters and Calculations
Vertical velocity: vy=2gh=2×9.8×19.6=384.16≈19.6m/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: Motion Parameters and Calculations
Vertical velocity: vy=2gh=2×9.8×19.6=384.16≈19.6m/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: Acceleration and Velocity Analysis - Part 8
Vertical velocity: vy=2gh=2×9.8×44.1=863.28≈29.38m/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 25 m/s 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
Vertical motion: y=12gt2, where y=−80m. 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 s 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
Time to fall: h=12gt2⇒24.5=12×9.8×t2⇒24.5=4.9t2⇒t2=5⇒t=5≈2.24s. 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 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
Time to fall: y = (1/2) g t², -80 = (1/2) × 10 × t² ⇒ t² = 16 ⇒ t = 4 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 48 m as the result, so option D is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Mixed Concepts
Horizontal distance x=vxt, so t=xvx=248=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 45 m as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Mixed Concepts
Vertical velocity: vy=2gh=2×10×12.5=250≈15.81m/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 15 m/s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Mixed Concepts
Ball’s speed relative to ground = 5+15=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 1 s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Uniform Acceleration and Graphical Analysis
Ball’s speed relative to ground = 6+10=16m/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 1 s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations
