If a ball is thrown straight upward with an initial velocity of v, it will reach a height of h.

step 1 So let's take our positive y -axis to be upward. And let's start with the kinematic equation tha

If a ball is thrown straight upward with an initial velocity...

Key Concepts

Scaling Relationships in Projectile Motion

The maximum height achieved by a projectile is proportional to the square of its initial velocity. This quadratic relationship implies that if the initial speed is doubled, the maximum height increases by a factor of four. Understanding this scaling behavior is essential for predicting changes in projectile behavior when initial conditions are altered.

Kinematic Equations

Kinematic equations are used to model the motion of objects under constant acceleration, linking initial velocity, time, acceleration, and displacement. They are crucial in analyzing projectile motion problems such as calculating the maximum height a projectile reaches when launched vertically.

Energy Conservation

The principle of energy conservation states that the total energy in an isolated system remains constant. In projectile motion, the conversion of kinetic energy at launch to gravitational potential energy at the maximum height provides an alternative approach to determining the height reached by the projectile.

Transcript

00:01 So let's take our positive y -axis to be upward.

00:08 And let's start with the kinematic equation that connects the initial velocities with the final velocity and the height reached.

00:21 So this is the equation 2 times acceleration times.

00:33 So let's take our positive y -axis to be upward.

00:40 And let's start with the kinematic equation that connects the displacement in y direction.

00:54 So let me just write it in a nicer way.

00:59 So a y means acceleration in the vertical direction and this is the amount of distance traveled in the vertical direction.

01:12 Now this is obviously the height, edge and acceleration velocities with the final velocity and the height reached.

01:26 So this is the equation 2 times acceleration times in vertical direction is going to be g.

01:36 But since it has a direction downward, therefore we are going to use negative g because we are also considering the sign here.

01:46 So this will give us v .f square to be equal to v .i.

01:50 Square minus plus twice of g h and we solve this for height so that will give us vf square minus v i square over 2g or the or at the highest point the object comes to rest for an instant and we can say that at that point the velocity is zero so basically our final velocity is 0 at the maximum height.

02:23 So this gives us the height to be negative v i square over 2g.

02:33 In vertical direction is going to be g but since it has a direction downward therefore we are going to use negative g because we are also considering the sign here.

02:46 So this will give us v.

02:48 Square to be equal to v i square minus plus twice of g h and we solve this for height so that will give us vf square minus vi i square over 2g now at the maximum height or the or at the highest point the object comes to rest for an instant and we can say that at that point the velocity is zero so basically our final velocity is 0 at the maximum height.

03:26 So this gives us the height to be negative v i squared over 2g.

03:40 Now doubling the initial speed of the ball will increase the maximum height by 4...

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