Question

A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

With upward as positive, \(v=u-gt\): a single straight line of constant negative slope that crosses zero at the top and continues into negative values.

Answer 1

The Correct Option is D

Net displacement is \(ut-\frac{1}{2}gt^2\), the area under the \(v\)-\(t\) graph. It must vanish at \(t=\frac{2u}{g}\), when the body is back at the starting point, so the areas above and below the axis must be equal, i.e. the velocity at \(t=\frac{2u}{g}\) is \(-u\) with the same constant slope \(-g\).
Only a single straight line from \(+u\) through zero to \(-u\) (graph 4) satisfies constant acceleration \(-g\) with this symmetry. The other graphs have a change of slope (1, 3) or a non-negative velocity (2), which would mean a different acceleration.
Answer: (4) the straight line with constant negative slope crossing the time axis