A body is projected with a speed u at an angle @ with the horizontal. The speed of the body when it is at the highest point on its trajectory is sqr.rt(2/5) times its speed at half the maximum height. The value of @ is?

To solve the problem of finding the angle of projection, we need to analyze the motion of the projectile at different points in its trajectory. The key points to consider are the speed at the highest point and the speed at half the maximum height. Let's break this down step by step.

Understanding Projectile Motion

When a body is projected at an angle, its motion can be divided into two components: horizontal and vertical. The initial speed \( u \) can be resolved into:

• Horizontal component: \( u_x = u \cos(\theta) \)
• Vertical component: \( u_y = u \sin(\theta) \)

Speed at the Highest Point

At the highest point of its trajectory, the vertical component of the velocity becomes zero. Therefore, the speed of the body at this point is solely due to the horizontal component:

Speed at highest point: \( v_{max} = u \cos(\theta) \)

Speed at Half the Maximum Height

To find the speed at half the maximum height, we first need to determine the maximum height \( H \). The maximum height can be calculated using the formula:

Maximum height: \( H = \frac{u^2 \sin^2(\theta)}{2g} \)

Half of the maximum height is \( \frac{H}{2} = \frac{u^2 \sin^2(\theta)}{4g} \).

At this height, the vertical component of the velocity can be found using the equation of motion:

Vertical velocity at half height: \( v_y = u \sin(\theta) - gt \)

To find the time \( t \) taken to reach half the maximum height, we can use the formula:

Time to reach maximum height: \( t_{max} = \frac{u \sin(\theta)}{g} \)

Since half the height is reached at half the time, we have:

Time to reach half height: \( t = \frac{u \sin(\theta)}{2g} \)

Substituting this time into the vertical velocity equation gives:

Vertical velocity at half height: \( v_y = u \sin(\theta) - g\left(\frac{u \sin(\theta)}{2g}\right) = \frac{u \sin(\theta)}{2} \)

Now, the speed at half the maximum height can be calculated as:

Speed at half height: \( v_{half} = \sqrt{(u \cos(\theta))^2 + \left(\frac{u \sin(\theta)}{2}\right)^2} \)

Expanding this gives:

\( v_{half} = \sqrt{u^2 \cos^2(\theta) + \frac{u^2 \sin^2(\theta)}{4}} = u \sqrt{\cos^2(\theta) + \frac{\sin^2(\theta)}{4}} \)

Setting Up the Equation

According to the problem, the speed at the highest point is \( \sqrt{\frac{2}{5}} \) times the speed at half the maximum height:

Equation: \( u \cos(\theta) = \sqrt{\frac{2}{5}} \cdot u \sqrt{\cos^2(\theta) + \frac{\sin^2(\theta)}{4}} \)

We can simplify this equation by canceling \( u \) (assuming \( u \neq 0 \)):

Final Equation: \( \cos(\theta) = \sqrt{\frac{2}{5}} \cdot \sqrt{\cos^2(\theta) + \frac{\sin^2(\theta)}{4}} \)

Solving for the Angle

Squaring both sides leads to:

\( \cos^2(\theta) = \frac{2}{5} \left(\cos^2(\theta) + \frac{\sin^2(\theta)}{4}\right) \)

Expanding and rearranging gives:

\( 5\cos^2(\theta) = 2\cos^2(\theta) + \frac{2\sin^2(\theta)}{4} \)

Substituting \( \sin^2(\theta) = 1 - \cos^2(\theta) \) leads to:

\( 5\cos^2(\theta) = 2\cos^2(\theta) + \frac{2(1 - \cos^2(\theta))}{4} \)

Solving this equation will yield the value of \( \theta \). After simplification, we find:

\( \theta = \tan^{-1}\left(\frac{1}{2}\right) \)

Thus, the angle of projection \( \theta \) is approximately \( 26.57^\circ \) or \( \frac{\pi}{6} \) radians.