A body projected vertically with a velocity 'u' from the ground. Its velocity
(a)At half of maximum height u/2
b) At `3//4^(th)` of maximum height `u/sqrt2`
(c) At `1//3^(rd)` of maximum height `sqrt(2/3)u`
(d) At `1//4^(th)` of maximum height `sqrt3/2u`

Options

A. a and b correct
B. b is correct
C. c and d correct
D. a is correct

Answer

To solve the problem of finding the velocity of a body projected vertically at different heights, we will use the kinematic equations of motion. The key points to consider are:

1. The maximum height \( H \) reached by the body when projected with an initial velocity \( u \) can be calculated using the formula:
\[ H = \frac{u^2}{2g} \]
where \( g \) is the acceleration due to gravity.

2. The velocity \( v \) of the body at any height \( h \) can be determined using the equation:
\[ v^2 = u^2 - 2gh \]

Now, we will calculate the velocity at the specified heights.

### Step 1: Calculate the maximum height \( H \)
Using the formula for maximum height:
\[ H = \frac{u^2}{2g} \]

### Step 2: Calculate the velocity at half of the maximum height \( \frac{H}{2} \)
Substituting \( h = \frac{H}{2} \) into the velocity equation:
\[ h = \frac{H}{2} = \frac{u^2}{4g} \]
Now substituting this value into the velocity equation:
\[ v^2 = u^2 - 2g\left(\frac{u^2}{4g}\right) \]
\[ v^2 = u^2 - \frac{u^2}{2} = \frac{u^2}{2} \]
Taking the square root:
\[ v = \frac{u}{\sqrt{2}} \]

### Step 3: Calculate the velocity at three-fourths of the maximum height \( \frac{3H}{4} \)
Substituting \( h = \frac{3H}{4} \):
\[ h = \frac{3H}{4} = \frac{3u^2}{8g} \]
Now substituting this value into the velocity equation:
\[ v^2 = u^2 - 2g\left(\frac{3u^2}{8g}\right) \]
\[ v^2 = u^2 - \frac{3u^2}{4} = \frac{u^2}{4} \]
Taking the square root:
\[ v = \frac{u}{2} \]

### Step 4: Calculate the velocity at one-third of the maximum height \( \frac{H}{3} \)
Substituting \( h = \frac{H}{3} \):
\[ h = \frac{H}{3} = \frac{u^2}{6g} \]
Now substituting this value into the velocity equation:
\[ v^2 = u^2 - 2g\left(\frac{u^2}{6g}\right) \]
\[ v^2 = u^2 - \frac{u^2}{3} = \frac{2u^2}{3} \]
Taking the square root:
\[ v = u\sqrt{\frac{2}{3}} = \frac{u\sqrt{2}}{\sqrt{3}} \]

### Step 5: Calculate the velocity at one-fourth of the maximum height \( \frac{H}{4} \)
Substituting \( h = \frac{H}{4} \):
\[ h = \frac{H}{4} = \frac{u^2}{8g} \]
Now substituting this value into the velocity equation:
\[ v^2 = u^2 - 2g\left(\frac{u^2}{8g}\right) \]
\[ v^2 = u^2 - \frac{u^2}{4} = \frac{3u^2}{4} \]
Taking the square root:
\[ v = \frac{\sqrt{3}}{2}u \]

### Summary of Results:
- At half of maximum height \( \frac{H}{2} \): \( v = \frac{u}{\sqrt{2}} \)
- At three-fourths of maximum height \( \frac{3H}{4} \): \( v = \frac{u}{2} \)
- At one-third of maximum height \( \frac{H}{3} \): \( v = \frac{u\sqrt{2}}{\sqrt{3}} \)
- At one-fourth of maximum height \( \frac{H}{4} \): \( v = \frac{\sqrt{3}}{2}u \)