If an object is thrown vertically up with a velocity of 19.6 m s^-1, it strikes the ground aft

If an object is thrown vertically up with a velocity of 19.6...

Key Concepts

Kinematic Equations

These are mathematical formulas that describe the motion of objects under constant acceleration. They relate displacement, initial velocity, final velocity, time, and acceleration, and are fundamental for solving problems involving projectile motion and free-fall.

Time of Flight

This concept refers to the total time an object spends in the air from the moment it is projected until it returns to its original level. It is typically determined using the kinematic equations, especially in scenarios where the object is subject only to gravity.

Acceleration due to Gravity

This is the constant acceleration imparted to objects due to the Earth's gravitational pull. In many physics problems, it is approximated as 9.8 m/s² and plays a crucial role in determining the motion of projectiles and freely falling bodies.

Projectile Motion

This refers to the motion of an object that is launched into the air and moves under the influence of gravity alone, neglecting air resistance. The analysis of projectile motion involves breaking the motion into horizontal and vertical components and applying kinematic equations to each.

Transcript

00:01 An object is projected upwards with an initial velocity of v0 meters per second from a point s0 meters above the ground.

00:11 We want to show that the velocity squared is equal to v0 squared minus 19 .6 times the precision s of t minus s0.

00:24 So this is a case of projectile motion with constant acceleration.

00:47 And this projectile motion is only in the y direction.

00:51 So i'm going to neglect all these subscripts for y because we're really dealing with a one -dimensional problem.

00:58 So our acceleration a is constant and is equal to minus the gravitational acceleration g.

01:10 And if i wanted to find the velocity now, what i could do is i could take the acceleration and integrate, giving us now that our velocity will be equal to minus gt plus segment integration constant c.

01:40 And using our initial condition that when t is equal to zero, we have an initial velocity of v0.

01:53 We find that c must therefore be equal to v0, implying that our velocity with constant acceleration along the y -axis will be equal to minus gt plus v0.

02:23 Now let's integrate our expression once more to find the position.

02:28 So s of t will be equal to the integral of v of t dt, giving us minus g2 squared over 2 plus v0 t plus another integration constant c.

02:53 And again, we're going to use our initial conditions to find c.

02:57 So when t is equal to zero, the position s is equal to s0, implying that our integration constant c this time will be equal to s0...

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