10 public questions tagged with this topic.
Time to max height: v=v0+at, 0=28−10t⇒t=2.8s. 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 2.8 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
Displacement y=−15m, −15=45t−5t2⇒5t2−45t−15=0⇒t2−9t−3=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 9 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
Time to reach max height: v=v0+at, 0=10−10t⇒t=1s. 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
Time to max height: v=v0+at, 0=50−10t⇒t=5s. 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 10 s as the result, so option C is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Uniform Acceleration and Graphical Analysis
At max height v=0. Using v²=u²-2gh from NCERT kinematics, h=u²/2g. With u=30 m/s and g=10 m/s², h=45.0 m. This matches option A (45 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: Uniform Acceleration and Graphical Analysis
At max height v=0. Using v²=u²-2gh from NCERT kinematics, h=u²/2g. With u=35 m/s and g=10 m/s², h=61.2 m. This matches option A (61.2 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: Uniform Acceleration and Graphical Analysis
Time to max height: v=v0+at, 0=24−10t⇒t=2.4s. 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 2.4 s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion in a Straight Line - Advanced Problems
Time to reach max height: v=v0+at, 0=30−10t⇒t=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 D is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion in a Straight Line - Advanced Problems
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
At max height v=0. Using v²=u²-2gh from NCERT kinematics, h=u²/2g. With u=12 m/s and g=10 m/s², h=7.2 m. This matches option A (7.2 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: Kinematic Equations and Uniformly Accelerated Motion
