From the top of a tower 19.6 m high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of ${45}^o$ with the horizontal, then the initial velocity of the ball is

Given that, height of tower h = 19.6m
 
Angle of line joining from point of projectile to the point of ground with horizontal,
$\theta ={45}^o$ 
In $\Delta AOB,\tan \theta =\frac{h}{R}$  
$\tan {45}^o=\frac{h}{R}\Rightarrow 1=\frac{h}{R}$ 
$\Rightarrow R=h=19.6m$ 
Where, R be the horizontal distance.
We know that, time taken to reach from point A to B, $t=\sqrt{\frac{2h}{g}}$
By substituting the values, we get $t=\sqrt{\frac{2\times 19.6}{9.8}}=2s$
Now, Horizontal distance = Horizontal velocity x Time $\Rightarrow R=ut$
19.6 = u x 2
[By substituting the value of R and t, we get]
$\therefore$ Initial velocity, u = 9.8 m/s