Show that the maximum height and Range of projectile are $\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}}\text{ and }\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}$.

a) Maximum Height of Projectile

The maximum height of a projectile is the highest vertical point reached by the projectile during its motion. At this point, the vertical component of velocity becomes zero.

Using the kinematic equation:

    v² - u² = 2as

Here:

• v = 0 (since the vertical velocity is zero at the maximum height).
• u = u sin(θ) (initial vertical velocity component).
• a = -g (acceleration due to gravity acting downwards).

Substitute the values:

    0² - (u sin(θ))² = -2g H_max

Simplify the equation:

    H_max = (u² sin²(θ)) / (2g)

Thus, the maximum height of projectile is given by:

H_max = u² sin²(θ) / 2g

b) Range of a Projectile

The range of a projectile refers to the total horizontal distance covered during its motion before returning to the same vertical level.

The formula for range (R) is derived using the horizontal velocity and the total time of flight:

    R = (Horizontal Velocity) × (Time of Flight)

The horizontal velocity is u cos(θ), and the total time of flight is given by 2u sin(θ) / g. Substituting these values:

    R = (u cos(θ)) × (2u sin(θ) / g)

Simplify the expression:

    R = (2u² sin(θ) cos(θ)) / g

Using the trigonometric identity sin(2θ) = 2 sin(θ) cos(θ), the equation becomes:

    R = (u² sin(2θ)) / g

Therefore, the range of the projectile is:

R = u² sin(2θ) / g

Conclusion

In summary, the maximum height of a projectile depends on the square of its initial velocity and the sine of the angle of projection. It is given by the formula:

H_max = u² sin²(θ) / 2g

The range of the projectile is determined by both horizontal and vertical components of velocity and is calculated as:

R = u² sin(2θ) / g

Understanding the concepts of maximum height of projectile and range is crucial for solving problems in projectile motion.