Question

An object thrown vertically up from the ground passes the height $5m$ twice in an interval of $10s$. The maximum height attained by it from the point of the projection is:

Answer 1

Change in velocity of the object $v-(-v)=2v$ {Velocity of ascent =velocity of descent}

$\frac{2v}{g}=t$

$\frac{2v}{9.8}=10$

$v=49m/s$

Height reach(t=5s,v=49 m/s),

$h=ut-\frac{1}{2}g{t}^2$

$h=49\times 5-\frac{1}{2}\times 9.8\times 5\times 5$

$h=245-61.25$

$h=183.75m$

Maximum height $=183.75+55=238.75$