14 public questions tagged with this topic.
Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields 30 m/s. Hence option D satisfies projectile formulas.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Relative Motion and Plane Motion
Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields 12.8 m. Hence option A satisfies projectile formulas.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Motion in a Plane and Projectile Motion
Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields 8 m. Hence option A satisfies projectile formulas.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Motion in a Plane and Projectile Motion
Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields Initial vertical velocity component. Hence option D satisfies projectile formulas.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Motion in a Plane and Projectile Motion
Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields 10 m. Hence option A satisfies projectile formulas.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Motion in a Plane and Projectile Motion
Velocity v=v₀+at. Starting from rest, v=at. With a=(ax i+ay j), magnitude |v|=t√(ax²+ay²) per NCERT vector addition. Substituting ax, ay, t gives |v| equal to 0 m/s. Hence option D is correct.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Motion in a Plane and Projectile Motion
Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields 5 m. Hence option A satisfies projectile formulas.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Motion in a Plane and Projectile Motion
Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields It decreases. Hence option B satisfies projectile formulas.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Motion in a Plane and Projectile Motion
At max height v=0. Using v²=u²-2gh from NCERT kinematics, h=u²/2g. With u=28 m/s and g=10 m/s², h=39.2 m. This matches option A (20 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: Motion Parameters and Calculations
At max height v=0. Using v²=u²-2gh from NCERT kinematics, h=u²/2g. With u=20 m/s and g=10 m/s², h=20.0 m. This matches option A (5 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: Acceleration and Velocity Analysis - Part 8
At max height v=0. Using v²=u²-2gh from NCERT kinematics, h=u²/2g. With u=20 m/s and g=10 m/s², h=20.0 m. This matches option A (2 s). Other options do not satisfy v²=u²+2as with correct signs.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Mixed Concepts
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
