A body is projected at an angle other than $90^{\circ}$ with the horizontal with same velocity. If the time of ascent of the body is ls, then the maximum height it can reach is (Take, $g=10 \mathrm{~ms}^{-2}$ )

Given, angle of projection, $\theta=90^{\circ}$ and time of ascent, $t=1$ s. Since, time of ascent of a projectile body is half of the time of height, $T=2 \times 1=2 \mathrm{~s}$ Equation of motion, $$ \mathrm{s}=u t+\frac{1}{2} g t^2 $$ where, $s=0$ for complete flight. and gravitational acceleration, $g=10 \mathrm{~m} / \mathrm{s}^2$ Putting the given values in above equation, we get $$ \begin{aligned} & 0=u \times 2-\frac{10}{2} \times 2^2 \\ & u=10 \mathrm{~m} / \mathrm{s} \end{aligned} $$ Now, the maximum height, $$ h=u t-\frac{10}{2} t^2 $$ where, $t=$ ls (time of ascent) $$ \begin{aligned} & h=10 \times 1-5 \times 1 \\ & h=5 \mathrm{~m} \end{aligned} $$ Hence, the correct option is (a).