91Ó°ÊÓ

In projectile motion, understanding the horizontal and vertical components is key to solving any related problems. Imagine you have a water cannon like in the exercise. It shoots water at a certain speed and angle. This speed, known as initial velocity, can be broken down into two parts, horizontal and vertical, using trigonometry.
The horizontal component (\( v_x \)) is responsible for the forward movement of the projectile. It can be found using the cosine of the angle with the formula: \( v_x = v_{initial} \cos{\theta} \). The vertical component (\( v_y \)), however, affects how high the projectile will go and is calculated with the sine of the angle: \( v_y = v_{initial} \sin{\theta} \).

• For example, with a velocity of 25 m/s at a 53-degree angle, the horizontal component is about 15 m/s.
• The vertical component is around 20 m/s.

These decomposed values help us understand and predict the motion path, breaking it into manageable sections.

Once we know our horizontal and vertical components, we can predict the path or trajectory of the projectile — in this case, the water stream. In physics, this path forms a parabolic shape due to gravity's pull.
The initial speed sends the projectile upward, but gravity slows it down until it stops for a brief moment at its highest point. Then, it pulls the projectile back down.
Visualizing the trajectory helps in planning where the water stream needs to land. The exercise above attempts to find out where to position the cannon, ensuring the water reaches a precise height. By sketching a diagram, the arc of the water can be seen clearly.

• This helps firefighters determine two points at height 10m where the stream will pass through: one on the way up, and another on the way down.

This parabolic path highlights how both angles of launch and speed influence the outcome of projectile motion.

Quadratic equations are mathematical representations often used to solve physics problems involving projectile motion. In the exercise, the vertical position is described by a quadratic equation, highlighting how changes in time affect the projectile's height.
The formula used is \( y = v_y t - \frac{1}{2}gt^2 \). We know our target height is 10 m. By substituting known values, we find \( 10 = 20t - 4.9t^2 \). This forms a quadratic equation, \( 4.9t^2 - 20t + 10 = 0 \), that can be solved using the quadratic formula.
Solving it allows us to find two different times when the projectile reaches this height:

• \( t_1 \approx 0.59 \ ext{ s} \)
• \( t_2 \approx 3.44 \ ext{ s} \)

These solutions correspond to two possible distances from the building where the cannon can be positioned. The quadratic nature of the equation reflects the reality of motion influenced by gravity, displaying both ascend and descend points within the trajectory.