A rock is thrown vertically upward from ground level at time `t=0`. At `t=1.5s` it passes the top of a tall tower, and 1.0 s later it reaches its maximum height. What is the height of the tower ?

Answer

To solve the problem of finding the height of the tower, we can follow these steps:

### Step 1: Determine the time to reach maximum height
The rock is thrown upwards and reaches its maximum height 1.0 seconds after it passes the top of the tower at t = 1.5 seconds. Therefore, the total time taken to reach the maximum height from the moment it was thrown is:
\[ t_{\text{total}} = 1.5 \, \text{s} + 1.0 \, \text{s} = 2.5 \, \text{s} \]

### Step 2: Use the kinematic equation to find the initial velocity
At maximum height, the final velocity \( v \) is 0 m/s. We can use the first equation of motion:
\[ v = u + at \]
Where:
- \( v = 0 \, \text{m/s} \) (final velocity at maximum height)
- \( u \) = initial velocity (which we need to find)
- \( a = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity, negative because it acts downwards)
- \( t = 2.5 \, \text{s} \) (time to reach maximum height)

Substituting the known values:
\[ 0 = u - 10 \times 2.5 \]
\[ 0 = u - 25 \]
Thus,
\[ u = 25 \, \text{m/s} \]

### Step 3: Calculate the height of the tower
Now we can find the height of the tower using the second equation of motion:
\[ h = ut + \frac{1}{2} a t^2 \]
Where:
- \( h \) = height of the tower
- \( u = 25 \, \text{m/s} \)
- \( a = -10 \, \text{m/s}^2 \)
- \( t = 1.5 \, \text{s} \) (time taken to reach the top of the tower)

Substituting the values:
\[ h = 25 \times 1.5 + \frac{1}{2} \times (-10) \times (1.5)^2 \]
Calculating each term:
1. \( 25 \times 1.5 = 37.5 \)
2. \( \frac{1}{2} \times (-10) \times (2.25) = -11.25 \)

Now, substituting these into the height equation:
\[ h = 37.5 - 11.25 \]
\[ h = 26.25 \, \text{m} \]

### Final Answer
The height of the tower is **26.25 meters**.

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