Motion Under Gravity

4.Distance covered by a body projected vertically up in the last one second of its upward journey = Distance covered by it in the 1^st second of its downward journey =

Sol. For a body falling downwards, we know that \( s = \frac{1} {2}gt^2 \)
Substituting t=1 gives us s= \( \frac{g} {2} \) (clearly, the distance travelled in 1^st second  in the downward journey is same as that of last second of upward journey)

5. Time taken by vertically projected up body to reach \(\frac{3} {4}\)th of maximum height =\( \frac{{t_a }} {2} \)
Sol. We know that, for a body projected vertically up 
     v=u-gt =>\( \frac{u} {2} \) =u-gt 
(since \( \frac{3} {4} \)th of maximum height, velocity =\( \frac{u} {2} \)) 
     \( gt = \frac{u} {2} \Rightarrow t = \frac{u} {{2g}} \Rightarrow t = \frac{1} {2} \times t_a \) 

6. For a Vertically thrown up body, maximum height \( \frac{1} {8}gT^2 \) where 'T' is the time of flight 
Sol. We know that time of flight T=\( \frac{{2u}} {g} \Rightarrow u = \frac{{gT}} {2} \)  
          Maximum height H = \( \frac{{u^2 }} {{2g}} = \frac{{\left( {\frac{{gT}} {2}} \right)^2 }} {{2g}} = \frac{1} {8}gT^2 \)
Similarly the same formula is applicable even in the case of a projectile. and stone after 't' sec is s =\( \frac{1} {2}(g + a)t^2 \)