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When dealing with projectile motion, kinematic equations are essential for understanding how an object moves through space. They help us calculate different aspects of motion like position, velocity, and acceleration. There are several key kinematic equations used often:

• \( v = v_0 + at \)Where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is acceleration, and \(t\) is time.
• \( s = s_0 + v_0 t + \frac{1}{2}at^2 \)This equation finds the position \(s\) at any time \(t\), starting from the position \(s_0\) with initial velocity \(v_0\) and constant acceleration \(a\).
• \( v^2 = v_0^2 + 2a(s - s_0) \)This relates the final velocity \(v\) with initial velocity \(v_0\), and displacement \(s - s_0\), and acceleration \(a\).

In projectile motion problems, horizontal and vertical motions are treated separately. This is due to gravity only affecting the vertical motion. Make sure to use the right equation for the part of motion you need to analyze.

Projectile motion often starts with an initial velocity given at an angle. To analyze it, split the initial velocity into horizontal and vertical components. This is done using trigonometric functions:

• For the horizontal component: \( v_{0x} = v_0 \cos(\theta) \)
• For the vertical component: \( v_{0y} = v_0 \sin(\theta) \)

Here, \(v_0\) is the initial speed, and \(\theta\) is the angle of projection with the horizontal. These components help you understand how much of the initial speed affects the horizontal and vertical movement. They are critical for further calculations like range or maximum height.
In the given problem, the initial speed was 30 m/s, projected at 53°. By breaking it down using the equations above, you can calculate how fast the rock initially moves both horizontally and vertically.

The range of a projectile is the horizontal distance it covers during its motion until hitting the ground. To find this, you first need to identify the total time of flight, which is the sum of the time it takes to reach maximum height, and the time taken to fall to the ground. This total time is crucial because the horizontal velocity remains constant in absence of air resistance.
To calculate the range \(x_R\), use:

• \( x_R = v_{0x} T \)

Here, \(v_{0x}\) is the horizontal component of the initial velocity, and \(T\) is the total time of flight.
In scenarios like the provided exercise, using the vertical motion equations can help find \(T\), which involves solving the kinematic equation for vertical displacement. Once \(T\) is determined, multiplying it by the horizontal component of velocity \(v_{0x}\) provides the range.

Calculating the maximum height of a projectile involves examining the highest vertical point reached before descending. At this point, the vertical velocity component is zero. To find this height, use:

• \( v_y^2 = v_{0y}^2 - 2g(h_{max} - h_0) \)

Set \(v_y = 0\) at maximum height to solve for \(h_{max}\), the vertical height relative to the starting point.
Another approach is using the time to rise to this peak. The time \(t\) at maximum height can be calculated from the initial vertical velocity using:

• \( 0 = v_{0y} - gt \)

Solving this gives the time to reach maximum height, which can then be used to find the maximum height itself through:

• \( h = y_0 + v_{0y}t - \frac{1}{2}gt^2 \)

Here, \(h_0\) is the initial height above the reference point the projectile was launched from, which in the problem was given as 100 m.