A particle is thrown with velocity 9u) making an angle ` theta` with the vertical. It just crosses the top of two poles each height (h) after ` 1 s` and ` 3 s` respectively . Fing the maximum hight of projecile. G= 9.8 ms//s^2`.

Explanation

Let the projectile go from ` O` to ` P` in time ` t-1 (=1 s) and from ` 0 to Q ` time `t_2 9=3s)`
Intital vetical velocity of particle at ` O=u cos theta`.
Taking vetical upward motin of particle from (O) to (P) and (O) to (Q), we have
.
` h = u cos theta t_1 - 1/2 gt_1 ^2 = u cos theta t-2 - 1/2 gt_1^2`
or ` u cos theta xx 1 - 1/2 g xx 1^2 = u cos theta xx 3 - 1/2 g 3^2` ltbRgt or ` u cos theta (3-1) = g/2 = g/2 (9- 1) = ( 9.8)/2 xx 8 = 4.9 xx 8`
or ` u cos theta = ( 4.9 xx 8)/2 = 4. 9 xx 4 = 19 .6 m//s`.
Max height, `H= (u^2 cos^2 theta) /( 2 g) = (( 19.6)^2)/( 2 xx 9.6) = 19 .6 m`.