A body is projected vertically upwards with some velocity u. It is at the same point at t = 6 s as it was at t = 4 s. Find the value of u.

$At{t}_1=4sand{t}_2=6s,bodyisatheighth$

So $h=u{t}_1-\left(\frac{1}{2}\right)g{t}_1^2$

$andh=u{t}_2-\left(\frac{1}{2}\right)g{t}_2^2$

Solve to get $u=\frac{1}{2}g(t_1+t_2)$

$=\frac{1}{2}\times 10(4+6)=50m/s$

Alternate: Time of flight: T = $t_1+t_2$

$\frac{2u}{g}=T\Rightarrow \frac{2u}{10}=10\Rightarrow u=50m/s$