A particle is projected vertically upwards and it reaches the maximum height H in time T seconds. The height of the particle at any time t will be-

Options

A. `g(t-T)^2`
B. `H-1/2 g (t-T)^2`
C. `1/2g(t-T)^2`
D. `H-g (t-T)^2`

Answer

To find the height of a particle projected vertically upwards at any time \( t \), we can use the equations of motion. Let's break down the solution step by step.

### Step 1: Understand the motion of the particle
When a particle is projected upwards, it will reach a maximum height \( H \) after a time \( T \). At this maximum height, the final velocity \( v \) of the particle is zero.

### Step 2: Use the first equation of motion
Using the first equation of motion:
\[ v = u + at \]
where:
- \( v = 0 \) (final velocity at maximum height)
- \( u \) is the initial velocity
- \( a = -g \) (acceleration due to gravity, acting downwards)

Substituting the values, we get:
\[ 0 = u - gT \]
This implies:
\[ u = gT \]

### Step 3: Use the second equation of motion to find maximum height
Now, we can use the second equation of motion to find the maximum height \( H \):
\[ H = uT - \frac{1}{2}gT^2 \]
Substituting \( u = gT \):
\[ H = gT \cdot T - \frac{1}{2}gT^2 \]
\[ H = gT^2 - \frac{1}{2}gT^2 = \frac{1}{2}gT^2 \]

### Step 4: Find the height at any time \( t \)
Now, to find the height \( h \) of the particle at any time \( t \):
Using the second equation of motion again:
\[ h = ut - \frac{1}{2}gt^2 \]
Substituting \( u = gT \):
\[ h = gT \cdot t - \frac{1}{2}gt^2 \]
\[ h = gTt - \frac{1}{2}gt^2 \]

### Step 5: Express height in terms of maximum height \( H \)
We know that \( H = \frac{1}{2}gT^2 \). We can express \( h \) in terms of \( H \):
\[ h = gTt - \frac{1}{2}gt^2 \]
\[ h = gTt - \frac{g}{H}t^2 \]
This can be rearranged to:
\[ h = H - \frac{g}{H}(T - t)^2 \]

### Final Result
Thus, the height of the particle at any time \( t \) is given by:
\[ h = H - \frac{g}{H}(T - t)^2 \]