Question

a ball is thrown vertically upwards from ground . is crosses a point at height of 25 m twice at interval of 4 sec . the ball was thrown with the velocity of

Answer 1

After reaching the point at 25 m height the ball travelled 4 second to reach the maximum height and come back to same point. So the to come back to the same point from maximum height it took 2 second. So the distance covered in the time is
$h_t=\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times 2^2=19.6\text{m}$
So the total height is
$h=19.6+25=44.6\text{m}$
Now if u was the initial velocity then maximum height is given by
$$\begin{gathered} u^2=2gh=2\times 9.8\times 44.6=874.16 \\ u=29.56 \end{gathered}$$
So the initial velocity was 29.56.