Question

In this question why time for motion above the tower is not considered for calculating time taken by the particle to hit the ground? Please help

Answer 1

Hey There!
The actual equations of Kinematics that we use are originally in vector form. So, the equation used here is actually:

$\vec{s} = \vec{u}t + \frac{1}{2} \vec{a} t^2$ 

Here, $\vec{s}$ refers to Displacement, and not the distance travelled.

As you know, to calculate displacement, you only need initial position and final position. 
Here, the ball starts from top of tower, and hits ground. Net displacement is only equal to the height of the tower. It does NOT depend on the path followed. 
That is why, we do not think of which path was followed. All factors are automatically covered by the equations. That is the beauty of Physics. 

So, actually, the motion above the tower is automatically considered. You only have to put the values properly with proper signs [positive or negative]. The equation will handle the rest. 

Hope it helps,
Purva
Brainly Community

Answer 2

Let us understand the given figure,

Total time includes time taken by the ball to go up above the point of projection and then come down to ground level. Origin is taken at the point of projection, height of tower. Keep track of directions as defined.