SSC JE Mechanical Engineering 11 Oct 2020 Shift 1 Papers

Part-I:
When the ball will reach the highest point then the final velocity will be zero.
Initial velocity = u m/sec, final velocity = 0 m/sec, acceleration = -g m/sec${}^2$
applying 1st equation of motion
v = u + at
$0=\mathit{u}-\mathit{g}{\mathit{t}}_1$
${\mathit{t}}_1=\frac{\mathit{u}}{\mathit{g}}$
Part-II:
Initial velocity will be zero as the ball is at the highest point.
applying 1st equation of motion
v = u + at
$3\mathit{u}=0+\mathit{g}{\mathit{t}}_2$
${\mathit{t}}_2=\frac{3\mathit{u}}{\mathit{g}}$
Therefore total time is:
$\mathit{t}={\mathit{t}}_1+{\mathit{t}}_2$
$\mathit{t}=\frac{\mathit{u}}{\mathit{g}}+\frac{3\mathit{u}}{\mathit{g}}=\frac{4\mathit{u}}{\mathit{g}}$