A body is projected at an angle other than 90^ 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 ~ms ⁻² )

Step-by-step solution

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