Question

a partical is thrown with velocity 'u' making on angle theeta with the vertical it just crossed the top of 2 poles each of hight 'h' after 1 second and 3 second respectively find the maximum height of projectile??​

Answer 1

Answer:

$t_H =\large\frac{v_0 y}{g} $ and

$H= \large\frac{v^2_0 y}{2g} =\large\frac{(gtH)^2}{2g} $$=\frac{1}{2} gt^2H$

Here $t_H =\large\frac{(1+3)}{2} $$=2 s$

Hence $H= \large\frac{1}{2}$$(9.8) (2)^2 =19.6\;m$

Hence B is the correct answer