A ball of mass $$m = 100\ \text{g}$$ is thrown vertically up with an initial velocity $$u = 98\ \text{m s}^{-1}$$. The ball goes up in the space and falls back to the ground. During its motion, the ball is observed to be at a certain height h at time $$t_1$$ and at time $$t_2$$. Also the speed of the ball is equal after time $$t_3$$ and $$t_4$$ from the start of its journey. With the given observation one can conclude that

This is a multiple-correct question, and the correct options are (a), (b) and (d). From $$h = ut - \frac{1}{2}gt^2$$, the two times at a given height satisfy $$t_1 + t_2 = \frac{2u}{g} = \frac{2 \times 98}{9.8} = 20$$ s and $$t_1t_2 = \frac{2h}{g}$$. Equal speeds occur at the same height, so $$t_3 + t_4 = 20$$ s too. The maximum height is $$\frac{u^2}{2g} = \frac{98^2}{19.6} = 490$$ m, not 980 m, and $$t_1t_2 = 50$$ gives $$h = \frac{9.8 \times 50}{2} = 245$$ m.