Near Earth’s surface and neglecting air resistance, every freely moving body has constant downward acceleration of magnitude g≈9.8 m s⁻², independent of its mass.

With upward positive: a=−g. For launch speed u from y₀:

v=u−gt
y=y₀+ut−½gt²
v²=u²−2g(y−y₀)

For return to the same level: time up=u/g, maximum rise H=u²/(2g), total time=2u/g, and return velocity=−u. These equalities assume no air resistance and identical launch/landing levels.

Example: u=19.6 m s⁻¹. Time to top=2 s; rise=19.6 m; total time to same level=4 s.

At the top, v=0 but a remains −g. A heavier object does not have greater free-fall acceleration in the ideal model.

Treat vertical motion as signed one-dimensional motion, taking +y upward and a=−g while air resistance is neglected. For an upward throw u>0, v=u−gt; at the top v=0 but acceleration stays −g. Rise above launch is H=u²/(2g), ascent time u/g, and return time to the same launch height 2u/g. A downward launch has u<0 under this convention. If final height differs from launch height, use y−y0=ut−½gt² rather than the symmetric-return time.

Worked example: A ball is thrown at 10 m s⁻¹ upward with g=10 m s⁻²: apex at 1 s, rise 5 m; after 2 s it returns to its launch height.

NCERT trap: Equal ascent/descent times apply only to equal heights with constant g and no drag.

NEET method: State the axis and substitute signed velocity and acceleration at each stage.

Reading estimate excludes time for the first three checks and the retake set.

Motion in a Straight Line

Check 1 of 3

Check 2 of 3

Check 3 of 3

Topic checkpoint