91Ó°ÊÓ

When we talk about projectile motion, it's essential to break it into two components: horizontal and vertical motion. These two motions are independent of each other.
The horizontal motion of a projectile, such as the ball in our exercise, happens at a constant velocity because there are no forces acting on it horizontally (ignoring air resistance). In our case, the initial horizontal velocity is given as 8.50 m/s.
Vertical motion, on the other hand, is affected by gravity. When the ball is thrown horizontally from the tree, it immediately begins to fall due to gravity. At the start, there is no vertical velocity component; it is only gravity pulling it downward, causing it to accelerate as it falls. As a result, the ball's velocity increases as it descends.
This separation of components helps us understand how a projectile moves:

• The horizontal motion helps us determine the distance traveled.
• The vertical motion helps us calculate how long it will take to hit the ground.

Gravity is a force that acts on all objects with mass, pulling them towards the center of the Earth. The standard acceleration due to gravity is symbolized as \( g \) and is approximately 9.81 m/s². This is crucial in calculating the time it takes for a projectile to hit the ground.
In our problem, the ball is thrown from a height of 12 m. Since it is thrown horizontally, we start with zero initial vertical velocity. Using the equation of motion \( y = \frac{1}{2} g t^2 \), we can solve for the time \( t \) it takes the ball to reach the ground.
Plugging in the provided values, we find the time to be approximately 1.56 seconds. This means that from the moment the ball is thrown, it will take about 1.56 seconds for it to fall and reach the dog's level, illustrating how gravity affects all objects, causing them to accelerate downwards uniformly.

Kinematics is the branch of physics that deals with motion without considering the forces that cause it. The equations of motion are handy tools that help us calculate various parameters such as displacement, velocity, acceleration, and time.
In solving projectile problems like ours, we often use several key kinematic equations. First, we use \( y = \frac{1}{2} g t^2 \) to find the time it takes for the projectile to fall to its target. Once we know the time, we can calculate how far it travels horizontally using \( x = v_x \cdot t \).
In our step-by-step solution, after determining the time (1.56 seconds), we used it to find the horizontal distance with the given constant horizontal speed (8.50 m/s). The final equation used illustrates how the kinematics equations provide all the necessary relations to fully describe the motion of projectiles in two dimensions.