Physics · Gravitation · NEET
You can only use v² = u² − 2gh when g is constant, which is true near the surface where the height is very small compared to Earth's radius R. When the body rises to a large height, g gets weaker with distance, so g is no longer constant. In that case you must use energy conservation with the real potential energy U = −GMm/r. Setting total energy at the surface equal to total energy at the top (where speed is zero) gives H = R/((2gR/u²) − 1).
It works when u is small, meaning the height H is much smaller than R (about a few km). Then g barely changes and mechanical energy near the surface gives (1/2)mu² = mgH, so H = u²/(2g). For NEET, use this only when the problem clearly involves small heights. If the speed is a large fraction of escape velocity, use the full formula H = Rk²/(1−k²), where u = k·v_escape.
Escape velocity is v_e = √(2gR). Writing u = k·v_e, the height is H = Rk²/(1−k²). As k approaches 1 (u approaches v_e), the denominator 1−k² approaches 0, so H tends to infinity. This means the body never comes back; it escapes Earth. This is exactly why v_e is called the escape speed.
No. The mass m cancels out in the energy equation because both kinetic energy (1/2 mu²) and potential energy (−GMm/r) contain m. So a heavy ball and a light ball thrown with the same speed reach the same maximum height (ignoring air resistance). The height depends only on u, g and R.
Use energy conservation from surface to the top. (1/2)mu² − GMm/R = −GMm/(R+H). Put u² = k²(2GM/R) since v_e² = 2GM/R. This gives (GM/R)(k²−1) = −GM/(R+H), so R+H = R/(1−k²) and H = Rk²/(1−k²). This is the exact NEET 2021 answer.
A particle of mass m is projected with velocity u = k·vₑ (k < 1) from the earth's surface (vₑ = escape velocity). The maximum height above the surface reached by the particle is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
With energy conservation, H = R/((2gR/u²) − 1), where u is the launch speed, R is Earth's radius and g is surface gravity. For small speeds this reduces to H = u²/(2g).
Writing u = k·v_e, the height is H = Rk²/(1−k²). When u equals v_e (k = 1), the denominator becomes zero and H becomes infinite, so the body never returns. That limiting speed is the escape velocity, v_e = √(2gR) ≈ 11.2 km/s for Earth.
No. Mass cancels in the energy equation, so all bodies thrown with the same speed reach the same height if air resistance is ignored.
Only when the height is very small compared to Earth's radius, so g stays nearly constant. If the launch speed is a large fraction of escape velocity, you must use the full energy-conservation formula instead.
