Vertical Projectile motion problems with solutions for Grade 12 PDF Download Projectile motion is a fundamental concept in physics, essential for understanding the behavior of objects launched into the air. In Grade 12 physics, students often encounter vertical projectile motion problems that challenge their understanding of kinematics and dynamics. In this blog post, we’ll delve into some typical problems encountered in Grade 12 physics curriculum related to vertical projectile motion, and provide step-by-step solutions to help students master this important topic.
Vertical Projectile motion
In Grade 12 Physical Sciences (Physics) we study the motion of objects that are moving upwards or downwards while experiencing a force due to gravity. This process is called projectile motion. We will only consider the case where objects move vertically upwards and/or downwards – meaning that there is no horizontal displacement of the object, only vertical displacement.
Main Vertical Projectile motion problems
When dealing with Projectile motion, students in Grade 12 should ensure they solve the following problems:
- Explain what is meant by a projectile, i.e. an object upon which the only force acting is the force of gravity.
- Use equations of motion to determine the position, velocity and displacement of a projectile at any given time.
- Sketch graphs for position vs time (x vs t), velocity vs time (v vs t) and acceleration vs time (a vs t) for each of the following: o A free-falling object.
- An object thrown vertically upwards
- An object thrown vertically downwar
- Bouncing objects (restricted to balls).
- For a given graph for x vs t, v vs t or a vs t, determine the following: o Position
- Velocity or acceleration at any time (t).
- For a given x vs t, v vs t or a vs t graph, describe the motion of the object for the following: o Bouncing.
- Thrown vertically upwards.
- Thrown vertically downwards.
Problem 1: A ball is thrown vertically upward with an initial velocity of 20 m/s from a height of 2 meters. Determine: a) The maximum height the ball reaches. b) The time taken for the ball to reach maximum height. c) The time taken for the ball to return to its initial height.
Solution: a) To find the maximum height, we can use the kinematic equation: v2=u2+2as Where: v = final velocity (0 m/s at maximum height), u = initial velocity (20 m/s), a = acceleration due to gravity (-9.8 m/s²), s = displacement (unknown).
Substituting the given values: 02=(20)2+2(−9.8)s 0=400−19.6s s=40019.6≈20.41 meters
So, the maximum height the ball reaches is approximately 20.41 meters.
b) To find the time taken to reach maximum height, we can use the equation: v=u+at Where: v = final velocity (0 m/s at maximum height), u = initial velocity (20 m/s), a = acceleration due to gravity (-9.8 m/s²), t = time taken (unknown).
Substituting the given values: 0=20−9.8t t=209.8≈2.04 seconds
So, the time taken for the ball to reach maximum height is approximately 2.04 seconds.
c) The time taken for the ball to return to its initial height is equal to the time taken to reach maximum height. Therefore, it’s also approximately 2.04 seconds.
Problem 2: A stone is dropped from the top of a cliff, which is 80 meters high. Determine: a) The time it takes for the stone to reach the ground. b) The velocity with which the stone hits the ground.
Solution: a) To find the time it takes for the stone to reach the ground, we can use the equation: s=ut+12at2 Where: s = displacement (80 meters), u = initial velocity (0 m/s for a dropped object), a = acceleration due to gravity (-9.8 m/s²), t = time taken (unknown).
Substituting the given values: 80=0+12(−9.8)t2 160=−9.8t2 t2=1609.8≈16.33 t≈16.33≈4.04 seconds
So, the time it takes for the stone to reach the ground is approximately 4.04 seconds.
b) To find the velocity with which the stone hits the ground, we can use the equation: v=u+at Where: v = final velocity (unknown), u = initial velocity (0 m/s for a dropped object), a = acceleration due to gravity (-9.8 m/s²), t = time taken (4.04 seconds).
Substituting the given values: v=0+(−9.8)(4.04) v≈−39.59 m/s
The negative sign indicates that the velocity is directed downward.