6 public questions tagged with this topic.
g(h) = g0(1+h/RE)2. 5.88 = 9.8(1+h/RE)2. (1+h/RE)2 = 1.667. 1+h/RE = 1.667≈1.291. h/RE = 0.291. h = 0.291×6.4×106≈1.86×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.9 × 10⁶ m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
g(h) = g0(1+h/RE)2. 7.84 = 9.8(1+h/RE)2. (1+h/RE)2 = 1.25. 1+h/RE = 1.25≈1.118. h/RE = 0.118. h = 0.118×6.4×106≈7.55×105m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.6 × 10⁵ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
g(h) = g0(1+h/RE)2. 7.35 = 9.8(1+h/RE)2. (1+h/RE)2 = 9.87.35≈1.333. 1+h/RE = 1.333≈1.154. h/RE = 0.154. h = 0.154×6.4×106≈9.86×105m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 9.9 × 10⁵ m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
At max height v=0. Using v²=u²-2gh from NCERT kinematics, h=u²/2g. With u=25 m/s and g=10 m/s², h=31.2 m. This matches option A (25 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: 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
Total time = t, last 1.2 s: 58.8=9.8t−4.9(t−1.2)2. 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 210 m as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Kinematic Equations and Uniformly Accelerated Motion
