Projectile Motion Basics

This section covers the kinematic equations for constant speed and acceleration.

Kinematic Equations

• Displacement is equal to velocity multiplied by time.

• Displacement is equal to the average velocity multiplied by time.

• Final velocity is equal to initial velocity plus acceleration multiplied by time.

• Square of final velocity is equal to square of initial velocity plus two times product of acceleration and displacement.

• Displacement can be calculated using the equation: displacement = initial position + (initial velocity x time) + (0.5 x acceleration x time^2).

Distance vs. Displacement

This section explains the difference between distance and displacement.

Distance vs. Displacement

• Distance and displacement are the same if an object moves in one direction without changing direction.

• If an object changes direction, distance and displacement are different.

Projectile Motion Example

This section provides an example of projectile motion with a ball being kicked off the ground.

Projectile Motion Example

• A ball is kicked off the ground with horizontal velocity v_x = 7 m/s and vertical velocity v_y = 30 m/s.

• One second later, vertical velocity decreases by 10 m/s due to gravitational acceleration, while horizontal velocity remains constant at 7 m/s.

• Three seconds later, vertical velocity becomes zero at the top of its trajectory, while horizontal motion continues at a constant rate of 7 m/s.

• A projectile is only under the influence of gravity, and air resistance is ignored.

• Speed is the absolute value of velocity, while velocity can be positive or negative.

Understanding Trajectories

In this section, we will learn about the three types of trajectories and their equations.

Types of Trajectories

• The first type is when a ball rolls off a cliff and falls down. The height of the cliff is represented by h, and the range is represented by r.

• The second type occurs when a ball is kicked off from the ground, goes up, and then comes back down. The maximum height of the trajectory is represented by h, and the horizontal distance is represented by r.

• The third type involves an object moving in a parabolic path due to gravity.

Equations for Trajectories

First Type

• To find the height (h), use the equation h = 1/2 * a * c^2.

• To calculate the range (r), use r = v_x * t.

Second Type

• To find the time it takes to go from point A to point B, use t = v * sin(theta)/g.

• To find the time it takes to go from point A to point C, simply double the value obtained for A to B due to symmetry.

• Maximum height (h) = v^2 * sin^2(theta)/2g

• Horizontal distance (r) = v^2 * sin(2*theta)/g

Third Type

• Use equations for projectile motion with constant acceleration.

Make sure you understand how each equation works and what each variable represents.

Trajectories and Equations of Motion

This section covers the three common trajectories and equations of motion used to solve problems related to them.

Vertical Trajectory

• The vertical speed was 20, and on the right side, the vertical velocity was negative 20.

• The final speed at point C is the same as the initial speed at point A.

• Keep in mind that the vertical velocity is positive at point A but negative at point C, but their magnitudes are equal.

Projectile Motion

• Projectile motion involves a ball kicked off a cliff or building at an angle, going up and then back down.

• All previously seen equations can be applied to this trajectory. Additionally, use y_final = y_initial + v_y_initial t + 1/2 a_y t^2 (where a_y is -9.8 m/s^2).

• To find time taken from A to C, use quadratic formula or add times taken from A to B and B to C.

Horizontal Trajectory

• To calculate speed just before hitting ground, use v_x constant and v_y changes.

• Find v_y final using v_y initial at point A. Then use equation for calculating final speed of ball just before it hits ground.

Problem Solving

This section covers how to solve problems related to projectile motion by applying equations of motion.

Problem Solving Steps

• Identify the type of trajectory.

• Make a list of known variables, including speed and time.

• Use equations to calculate height and horizontal distance traveled by the ball.

Example Problem

• A ball rolls horizontally off a cliff at 20 m/s. It takes 10 seconds for it to hit the ground. Calculate the height of the cliff and horizontal distance traveled by the ball.

• h = 4.9 x 100 = 490m

• Horizontal distance = v_x x t = 20 x 10 = 200m

Projectile Motion

In this section, we solve two problems related to projectile motion. The first problem involves finding the range of a projectile given its initial velocity and angle of launch. The second problem involves finding the time it takes for a ball to hit the ground when dropped from a certain height or thrown with an initial velocity.

Finding Range of a Projectile

• To find the range of a projectile, we use the equation R = (v^2 * sin(2θ))/g, where v is the initial velocity, θ is the angle of launch, and g is acceleration due to gravity.

• Using this equation, we can calculate that if a ball is launched at an angle of 45 degrees with an initial velocity of 20 m/s, it will travel 200 meters before hitting the ground.

Finding Time for Ball to Hit Ground

Dropped Ball

• To find the time it takes for a ball dropped from rest to hit the ground, we use h = (1/2)gt^2 where h is height and g is acceleration due to gravity.

• If a ball is dropped from a cliff that is 200 meters high, it will take approximately 6.39 seconds to hit the ground.

Thrown Ball

• To find the time it takes for a ball thrown straight down with an initial speed of 30 m/s to hit the ground, we use d_y = v_y0 t + (1/2)gt^2 where d_y is vertical displacement and v_y0 is initial vertical velocity.

• Solving this quadratic equation using the quadratic formula gives us two possible solutions: t = 10.2 seconds or t = -16.5 seconds. Since time cannot be negative, we discard the negative solution and conclude that it takes approximately 10.2 seconds for the ball to hit the ground.

Conclusion

In this section, we learned how to solve problems related to projectile motion by finding the range of a projectile and the time it takes for a ball to hit the ground when dropped or thrown with an initial velocity.

Solving for Time

In this section, the speaker solves for time using a given equation.

Solving for Time

• The equation is negative four times four point nine times negative eight hundred.

• This is equal to positive fifteen thousand six hundred and eighty.

• Adding 900 to 15,680 gives us 16,580.

• The square root of 16,580 is 128.76.

• There could be two answers: negative 30 minus 128.76 and plus 128.76. We know time can't be negative so we're just going to ignore the negative answer.

• If we take negative 30 plus 128.76 that's positive 98.76 and divided by 9.8 this will give us a t value of 10.08 seconds which is less than the other answer which was 12.78 because in the second example the ball was thrown towards the ground it's going to take a shorter time to get to the ground since it was given that initial speed and so it makes sense why it would be less and we could check the answer if you plug it into the original equation we need to make sure that we get zero point four point nine times ten point zero eight squared plus thirty times ten point zero eight that's eight hundred point twenty seven minus eight hundred that's approximately zero so this answer is correct