A ball 4 s after the instant it was thrown from the ground passes through a point P, and strikes the ground after 5 s from the instant it passes through the point P. Assuming acceleration due to gravity to be `9.8 m//s^(2)` find height of the point P above the ground.

Explanation

To find the height of point P above the ground, we can follow these steps:

### Step 1: Understand the problem
The ball is thrown from the ground and passes through point P after 4 seconds. It then strikes the ground 5 seconds later, meaning it takes a total of 9 seconds from the moment it was thrown to hit the ground.

### Step 2: Use the time of flight formula
The time of flight \( T \) for a projectile thrown vertically can be expressed as:
\[ T = \frac{2U_y}{g} \]
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