91Ó°ÊÓ

Horizontal velocity is a crucial component in the study of projectile motion. It refers to the constant speed at which a projectile travels horizontally. In the context of our exercise, the projectile's horizontal velocity, denoted as \(v_h\), was calculated under the influence of uniform acceleration. Consider a body starting from rest, which means its initial velocity \(u\) is zero. If it moves with an acceleration of \(3 \, \text{m/s}^2\) for \(0.5\) minutes, you first convert the time to seconds giving \(30\) seconds.

To calculate \(v_h\), use the equation of motion:

• \(v = u + at\)

Given \(u = 0\), \(a = 3 \, \text{m/s}^2\), and \(t = 30\, \text{s}\), the equation becomes:

• \(v_h = 0 + 3 \times 30 = 90 \, \text{m/s}\)

Horizontal velocity remains unaffected by forces along the vertical direction like gravity, making it constant throughout the projectile's motion.

The maximum height of a projectile is a measure of how high it rises from the point of projection before starting its descent. Determining this height offers insights into the vertical component of velocity at the initial stage. In this exercise, the maximum height is given as \(80 \, \text{m}\).

To find the vertical component of velocity \(v_v\), you can use the formula:

• \(H = \frac{v_v^2}{2g}\)

where \(H\) is the maximum height and \(g\) represents the acceleration due to gravity, approximately \(10 \, \text{m/s}^2\) on Earth. Solving it using the given height:

• \(80 = \frac{v_v^2}{2 \times 10}\)
• \(v_v^2 = 1600\)
• \(v_v = 40 \, \text{m/s}\)

This vertical velocity component helps in finding the angle of projection, ultimately influencing the projectile's trajectory.

The angle of projection is the initial angle at which a projectile is launched concerning the horizontal plane. This angle significantly affects both the range and the maximum height of the projectile.

In the problem, we determine the angle \(\theta\) using the relation between vertical velocity \(v_v\) and horizontal velocity \(v_h\):

• \(\tan \theta = \frac{v_v}{v_h}\)

Given \(v_v = 40\, \text{m/s}\) and \(v_h = 90\, \text{m/s}\), substituting the values results in:

• \(\tan \theta = \frac{40}{90} = \frac{4}{9}\)
• So, \(\theta = \tan^{-1}\left(\frac{4}{9}\right)\)

Such calculations are pivotal in predicting where and how the projectile will land, shedding light on designing trajectories in fields like sports and engineering.

Equations of motion are fundamental in physics, especially in analyzing projectile motion. These equations allow us to describe the motion of a projectile under uniform acceleration.

In the context of our exercise, we used the first equation of motion:

• \(v = u + at\)

This equation solved for horizontal velocity, establishing how the given acceleration over time contributes to the final speed. Additionally, for maximum height, the formula related to vertical motion was:

• \(H = \frac{v_v^2}{2g}\)

Each equation serves to unravel different aspects of projectile motion. The uniform motion equations, coupled with trigonometric identities, enable us to correlate the kinematic parameters, laying the groundwork for solving complex projectile problems efficiently.