A ball is thrown with an initial speed vi at an angle θi...

Key Concepts

Maximum Height

This refers to the highest point reached by the projectile during its flight. It is determined solely by the initial vertical component of the velocity and gravitational acceleration, using the kinematic equations to relate these quantities.

Time of Flight

The time of flight is the total time the projectile remains in motion before landing. It is influenced by the initial vertical velocity, the acceleration due to gravity, and the symmetry of the motion, and can be derived using kinematic relations.

Horizontal Range

The horizontal range is the total distance traveled by a projectile along the horizontal axis. It depends on both the initial speed and the angle of projection, and can be derived from the horizontal component of the motion combined with the time of flight.

Trigonometric Relations

Trigonometry is used to relate the components of the initial velocity to its magnitude and the angle of projection. Using sine, cosine, and tangent functions allows conversion between the overall speed and its horizontal and vertical components, which is essential in analyzing projectile motion.

Decomposition of Velocity

In projectile motion, the initial velocity is typically split into horizontal and vertical components using trigonometric functions. The horizontal component affects the range while the vertical component is responsible for the ascent, descent, and maximum height achieved by the projectile.

Projectile Motion

This concept involves the motion of an object thrown into the air, which then moves under the influence of gravity. The key assumption is that the only acceleration acting on the object is the constant acceleration due to gravity, and air resistance is neglected. This allows the motion to be analyzed separately in horizontal and vertical components.

Kinematic Equations for Uniform Acceleration

These equations describe the relationships between displacement, velocity, acceleration, and time in scenarios with constant acceleration. They are fundamental to solving problems in projectile motion, enabling the calculation of variables such as time of flight, maximum height, and displacement in horizontal and vertical directions.

Transcript

0:00 Hello.

00:02 In this problem, we have a ball that is thrown with an angle with the horizon.

00:09 Let's call it theta.

00:11 So the ball is thrown and of course it's make it's a projectile and it's thrown with angle theta and initial velocity vi.

00:22 Such that the horizontal range of this ball, let's call it r, while the maximum height that the ball can reach is r.

00:31 Over 6.

00:34 So first in part a, we're asked to calculate the time that the ball is in motion.

00:40 So let's first write these two formulas here.

00:45 We have the formula of the maximum height, which is v initial squared sine squared theta divided by 2g.

00:59 The second equation that i'm going to use here is the equation for the time of flight for a projectile, which is to v .i.

01:10 Sine theta divided by g.

01:17 So here, let's substitute with h equal to r over six, because that's the maximum height here.

01:25 So we know that r over six is equal to v .i squared sine squared theta divided by 2g, from here, we can get the value of v .i.

01:41 Squared .s .n.

01:43 Squared theta.

01:46 And it's equal to rg over 3.

01:51 Of course, taking the square root of this of both sides, we know that vi .s .a.

01:58 Is equal to the square root of rg over 3.

02:02 And let's box this useful piece of information here.

02:06 Then we can substitute by the value of v -i -sign -theta here in this equation in order to get the time of flight in terms of r and g as requested in the problem.

02:20 So now, t is equal to 2 over g, multiplied by v -i -syne -theta, which is the square root of r -g over 3.

02:36 Getting this term here inside the square root, we know that it's equal to 4rg divided by 3g squared.

02:52 This will cancel with the square here.

02:55 And we have the answer that the time of flight is 4 over 3 r over g.

03:04 So that's the answer for part a here.

03:10 In part b, we're asked to calculate the speed of the ball at its peak of the path.

03:19 So at the peak of the path, so here, for example, we know that the vertical component of the velocity is zero, and the only component of the velocity is the horizontal component.

03:35 So the velocity at the peak is equal to the horizontal component, which is equal to the horizontal component, the initial velocity times cosine theta.

03:46 So here our task is to find cosine theta or, in other words, to find theta itself.

03:52 And that's actually, we are asked to get this as well, so we have to do it anyway.

03:59 Okay, so let's write the formula again for the range of a projectile, which is v .i.

04:05 Squared sine 2 theta divided by g and the equation of maximum height, which is vi squared, sine squared theta, divided by 2g.

04:23 Here, let's substitute for the range.

04:26 The ranges are already, and the height, the maximum height, is r over 6.

04:32 So multiplying this equation by 6, we can make it equal to this equation.

04:38 Here.

04:39 So we know that v .i squared, that's equal to r, v .i squared, sine 2 theta over g is equal to 6, multiplied by the height, which is v .i squared, sine squared theta divided by 2g.

05:02 Here, g will cancel the g, and the i squared will cancel as well.

05:08 And we have this sign 2, theta, which we know that this is a trig identity, that sine 2 theta is equal to 2 .5 ,000, theta, cosine, theta.

05:24 So we will replace this here...

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