Grade 9 Science Projectile Motion - PDF

Certainly! Here is the conversion of the attached document into a markdown format # Grade 9 Science Quarter 4 Week 2 ## Projectile Motion Launched at an Angle --- ### Learning Objective * Investigate the relationship between the angle of release and the height and range of the projectile. ---...

Certainly! Here is the conversion of the attached document into a markdown format # Grade 9 Science Quarter 4 Week 2 ## Projectile Motion Launched at an Angle --- ### Learning Objective * Investigate the relationship between the angle of release and the height and range of the projectile. --- * Trajectory - the arc/curve-like motion-path undergone by a projectile. * Projectile - an object undergoing projectile motion. * The horizontal component of projectile motion has the acceleration equal to zero since the velocity is constant. * The vertical component of acceleration is constant which is acceleration due to gravity ($9.8 m/s^2$). --- The following points describe the trajectory of a projectile such as a ball thrown by a baseball player: * Angle - a numerical value in degrees (°) expressing the orientation of a projectile to be thrown/projected. * Vertical Velocity - vertical velocity responsible for the projectile to travel at a vertical distance. * Horizontal Velocity - horizontal velocity responsible for the projectile to travel at a horizontal distance. * Maximum Height - the maximum vertical distance from the earth's a projectile can reach. * Range - horizontal distance covered by a projectile Additional key points in the image relating to the trajectory of the baseball include: * As the projectile rises (point A to B) the vertical velocity ($V_y$) is decreasing, this is because the direction of gravity is opposite to the projectile motion. * As the projectile reaches the maximum height (point B) it momentarily stops causing a vertical velocity equal to zero ($V_y = 0$). * When it returns back to the ground (point B to C) it agrees to the direction of gravitational force hence vertical velocity is increasing. ## Variables involved in Projectile Launched at an Angle | HORIZONTAL COMPONENT (X) | VERTICAL COMPONENT (Y) | | :------------------------------ | :------------------------------------ | | $d_x =$ horizontal displacement | $d_y =$ vertical displacement | | $V_x =$ horizontal velocity | $V_y =$ vertical velocity | | $V_{ix} =$ horizontal initial velocity | $V_{iy} =$ vertical initial velocity | | $V_{fx} =$ horizontal final velocity | $V_{fy} =$ vertical final velocity | | $a_x =$ horizontal acceleration | $a_y =$ vertical acceleration acceleration | | $t_T =$ total time | $g =$ acceleration due to gravity = $9.8 m/s^2$ | --- ### FACTS ABOUT PROJECTILE LAUNCHED AT AN ANGLE 1. An object is projected from rest at an upward angle $\theta$. 2. Its initial velocity can be resolved into two components. 3. The horizontal velocity is constant due to gravity. $V_x = constant$ $V_x = V_{ix} = V_{fx}$ $a_x = 0$ 4. The amount of time the object takes to come to a stop at its highest point is the same amount of time it takes to return to where it was launched from. $t_{up}$ = $t_{down}$ 5. The initial velocity upward will be the same magnitude (opposite in direction) as the final velocity when it returns to its original height. at the maximum height: $V_f = 0$ $a_y = g = 9.8 \frac{m}{s^2}$ $V_f = -V_i$ --- ## Equations for Projectile Launched at an Angle | QUANTITIES | HORIZONTAL COMPONENT | VERTICAL COMPONENT | | -------------------------: | :----------------------- | :------------------------------- | | d | $dx = V_{ix} \cdot t$ | $dy = \frac{V_{iy}^2}{2g}$ | | | $dx = V_i \cdot cos\ \theta\cdot t$ | $dy = \frac{(V_i \cdot sin\ \theta)^2}{2g}$ | | Vi | $V_{ix} = V_i \cdot cos\ \theta$ | $V_{iy} = V_i \cdot sin\ \theta$ | | Vf | $V_{fx} = V_{ix}$ | $V_{fy} = -V_{iy}$ | | | $V_{ix} = V_i\ cos\ \theta$ | $V_{iy} = V_i \cdot sin\ \theta$ | | | $V_{fx} = V_i \cdot cos\ \theta$ | $V_{fy} = -V_i \cdot sin\ \theta$ | | a | $a_x = 0$ | $a_y = g = 9.8 \frac{m}{s^2}$ | | t | $t = \frac{dx}{V_{ix}}$ | $t_{up} = \frac{V_{iy}}{g}$ | | |$t = \frac{dx}{V_i \cdot cos\ \theta}$| $t_T = (2) \cdot ( \frac{V_{iy}}{g})$ | | | | $t_T = (2) \cdot ( \frac{V_i \cdot sin\ \theta}{g})$ | --- The following points describe visual trajectory representations for a projectile launched at varying degrees: * A soccer player is shown kicking a ball at $15°$, $30°$, $45°$, $60°$, and $75°$. * A cannon is shown launching a ball at $30°$, $45°$, and $60°$. * The maximum range is achieved if the projectile is fired at an angle of $45°$ with respect to the horizontal component. * An object launched at an angle of $30°$ will also share the same range if it is launched at $60°$. * The angles $30°$ and $60°$ are called complementary angles because they add up to $90°$. * As the angle of launch increases, the vertical displacement of the projectile will also increase. At the highest point, the vertical component of velocity is zero and the time to reach the maximum height is half of the total time of flight. ### Sample Problem: A baseball player leads off the game and hits along home run. The ball leaves the bat at an angle of 25° with a velocity of 30 m/s. a. What is the maximum height reached by the ball? b. What is the horizontal displacement (range) of the ball? Given: $V_i$ = $30 \frac{m}{s}$ $\theta = 25 °$ $g = 9.8 \frac{m}{s^2}$ Formula: $d_y = \frac{(V_i \cdot sin\ \theta)^2}{ 2g}$ a) What is the maximum height reached by the ball? $d_y = \frac{(30 \frac{m}{s} \cdot sin\ 25 °)^2}{ (2)(9.8 \frac{m}{s^2})}$ $d_y=\frac{160.745 ...\frac{ m^2}{ s^2}}{19.6 \frac{m}{ s^2}}$ $d_y= 8.20 m$ $d_y = 8.20 m$ b) What is the horizontal displacement (range) of the ball? $dx = Vi \cdot cos\ \theta \cdot t $ $t_T = (2) \cdot \frac{Vi \cdot sin\ \theta }{g}$ $t_T = (2) \cdot \frac{(30 \frac{m}{s}) \cdot (sin\ 25 °) }{9.8 \frac{m}{s^2}}$ $t_T = (2) \cdot \frac{12.678 \frac{m}{s}}{9.8 \frac{m}{s^2}}$ $t_T = 2.59 s$ $d_x = (30 \frac{m}{s})\cdot (cos\ 25°) (2.59 s)$ $d_x = 70.42 m$ --- THE END. Let me know if you have any other questions.