00:02 Good day, everyone.

00:15 My name is Mr. Bakare.

00:17 I'm your physics teacher today,

00:21 and the topic we'll be teaching today is projectiles.

00:28 Now, before we move on,

00:30 let's look at the objectives of the topic.

00:33 Number one, at the end of this topic,

00:37 today should be able to define projectile

00:41 and projectile motion.

00:43 Number two, give examples of projectile in spot

00:48 and warfare.

00:51 Number three, today should be able to define

00:54 one time of flight, maximum height and range of projector.

00:59 And number four, today should be able to state equations,

01:05 formula for the time of flight,

01:08 maximum height and range.

01:11 And number five, today should be able to state

01:14 the applications of projector.

01:16 And lastly, should be able to solve simple problems.

01:23 Now, what is a projectile?

01:25 It's a common experience that you'd rather love

01:28 to choose tones to bat on tree or through a stick

01:33 to a tree in order to block a mango from a mango tree.

01:38 This tone-trune into the air or stick

01:41 is regarded as a projectile.

01:44 Like we have in the diagram a child

01:48 trying to play football.

01:52 So let's define projectile motion.

01:55 Projectile motion is the curved motion

01:59 described by a falling object in the air.

02:03 Examples of body's exhibiting projectile motion

02:08 are one in spot, a trone javelin,

02:13 a trone discourse, a trone shot put,

02:20 a bouncing table tennis ball from the wall

02:24 and a fired missile.

02:25 All these examples can be found in spot.

02:29 Then in warfare, we have a trone bomb,

02:36 a trone spear, a fired bullet from a gun,

02:42 and a shot stone from a catapult.

02:46 As you can see, the diagram of a catapult there.

02:51 Let's now define the terms involved in a projector.

02:56 One, time or flight.

02:58 Time or flight.

03:00 If an object is projected from a velocity U,

03:04 we take U to represent initial velocity of projector.

03:10 And at an angle of inclination theta

03:15 with the horizontal, the vertical component of the velocity

03:20 which is U y and it's time to be U sin theta

03:25 while the horizontal component U x

03:28 is sent to be U cos theta.

03:30 Now in our previous lesson,

03:33 we talked about a vertical and horizontal component.

03:38 Now we say the vertical component is U sin theta

03:42 while the horizontal component U x is U cos theta.

03:49 Then at maximum height, the final velocity,

03:53 final vertical velocity of the projector is zero.

03:57 Just like we said in the last session

04:02 that when an object is projected vertically upward,

04:06 it will get to a particular height

04:08 that it will be able to move upward again.

04:10 It will stop momentarily before coming down.

04:14 And that is why the velocity is zero.

04:17 At that moment of stopping.

04:19 So we now have from our first equation of motion,

04:22 V equal to U plus GT.

04:25 We have, but because the object is going vertically upward,

04:30 which is against the force of gravity.

04:32 So we assign a negative sign to it,

04:36 which is V, V equal to U plus GT.

04:44 But because we are projecting against gravity.

04:48 Therefore, our equation will be V equal to U minus GT, right?

04:53 Now we said our velocity is zero at maximum height.

05:04 And we are projecting vertically upward.

05:06 That means we are resolving vertically,

05:09 which is U sin theta.

05:13 If we are resulting horizontally,

05:14 it's going to be U cos theta.

05:17 So U sin theta minus GT.

05:21 So if you make your T, the sub y of the formula.

05:26 So we have GT minus GT.

05:30 We cross equality sin to become plus GT

05:32 equal to U sin theta.

05:37 And T equals to U sin theta over G.

05:46 So this T is small at a T.

05:50 It's a time to reach the maximum height.

05:54 Time to reach the maximum height.

05:57 We have two times.

05:58 Time to reach maximum height.

06:00 And time to get to the projection plane.

06:05 Those are the two times.

06:06 So those two times is our time of flight.

06:13 It's our time of flight, which is capital letter T equal to 2T,

06:17 which is 2 times this T as U sin theta over G.

06:27 So we have T equal to 2U sin theta over G.

06:37 So that's our time of flight as it is.

06:45 If the time taking to reach the maximum height is T1,

06:48 and T2 is the time taking to return

06:50 to the plane of projection, so T1 equal to T2 equal to T.

06:54 So we have capital letter T equal to 2T, which is 2U sin theta over G.

07:02 So our capital letter T is the time required for projectile

07:06 to return to the plane of projection.

07:10 Now, what about the maximum height?

07:13 The maximum height is represented with capital letter H.

07:24 The highest vertical distance attained by projectile

07:29 as measured from the horizontal plane of projection

07:33 is the highest vertical distance attained.

07:36 So our H to find H, we make use of the second equation.

07:43 We have V squared equal to U squared minus 2G H.

07:51 It's T minus because we are projecting vertically upward.

07:55 Now, we want to make H subject of the formula for here.

07:58 And we said, mind you, that V is 0 at maximum height.

08:03 And vertically, U is U sin theta.

08:06 So why is substitute for that?

08:08 We have 0, 0 squared is 0 equal to U sin theta all squared minus 2G H.

08:21 So bringing to G H to this place, we have 2G H plus 2G H equal to this U squared sin squared

08:32 theta when you expand.

08:35 Then we have H equal to U squared sin squared theta over 2G.

08:46 So that is our equation for the maximum height.

08:52 What about the range?

08:55 The range of projection.

08:57 The range of projector is the horizontal distance

09:02 covered by a projector from the point of projection

09:06 to the point where the projector lands on the projection plane again.

09:11 Now, if the time taken to cover the distance is T.

09:15 Now, we are taking capital letter T because we

09:18 assumed that is the time of projection.

09:21 For instance, if someone is coming from London through flight,

09:27 we can call him and tell and ask him, hello, how was your flight?

09:31 Because he has not landed.

09:33 So until he lands before we can be asking such questions.

09:36 So that's why we are using capital letter T as our time of flight.

09:42 Now, we want to get the range.

09:45 Now, we are going to make use of the equation R which is H equal to U T plus R GT squared.

10:05 Now on the horizontal plane according to the definition of range, there is no gravity

10:11 on the horizontal.

10:12 So definitely our G is 0.

10:15 If G is 0, the whole of this is 0.

10:18 So we are left with R equal to U T capital letter T because it is the time of flight.

10:28 The object has landed.

10:30 So we now substitute the horizontal.

10:32 Now, we have R equal to U cos theta horizontal.

10:39 And times our capital letter T, I hope you remember, 2U sin theta over G times 2U sin theta

10:58 over over G.

11:01 So after a little bit of a trigonometry, we can say that R equal to U squared sin 2 theta

11:17 over G. So these are formula for the range of projectile.

11:28 Yeah, so R equal to U squared sin 2 theta over G.

11:34 When theta is 45 degrees, we say that range is maximum.

11:41 At 45 degrees, our range is maximum.

11:45 So 2 theta, that's 2 times 45, that's 90.

11:49 And sin 90 is 1.

11:52 So if sin 90 is 1, we have R equal to U squared over over G.

12:03 So hence, maximum horizontal range can be achieved when our theta or angle of projection

12:10 is 45 degrees.

12:14 Then let's look at the application of projectile.

12:17 Number one is used in warfare to hit enemy target, number two.

12:22 It is used in sports by throwing javelin discourse, et cetera, acts angle 45 degrees to obtain

12:29 a maximum range.

12:31 Now let's look at an example here to solve.

12:35 A missile is fired at a speed of 200 meters per second at an angle of 45 degrees.

12:42 Find, the height obtained by the missile, how long does it take in the air?

12:51 How long means time?

12:53 Then number three, the horizontal range.

12:56 Now the solution, the angle of projection is 45 degrees.

13:06 As our theta is 45 degrees, our velocity of projection is 200 meters per second.

13:16 Then G is 10 meters per second squared.

13:20 Then we are looking for the height, height equal to U squared sin squared theta over 2G.

13:33 That's the formula we derived the other time.

13:36 So by inserting our parameters in this formula, we have U squared U is 200 squared times sin

13:48 squared theta means the square of the sine of the angle, the angle is 45 degrees.

13:55 So sin 45, all squared, so we have sin 45 degrees, all squared, then over 2 times 10.

14:12 So at the end of the day, how 2, that's 4, there are 4, 1, 2, 3, 4 times 0.7, 0.7, 1, all squared

14:27 over 20, yeah.

14:31 So then we have 1000, 1000 meters as our height.

14:38 And we are looking for the time of flight, the time it's spent in A. That's the time of flight,

14:46 that's capital letter T, which is 2U sin theta over G. So our T, 2U sin theta over G equal

14:58 to 2 times 200 times sin 45 over 10.

15:07 At the end of the day, our time of flight, time of flight we have 28.28 seconds.

15:21 The range, the range, the range, bearing the mind that R equal to U squared sin 2 theta

15:34 over G, then by inserting the parameters, we have R is 200 squared times sin 2 theta means

15:45 2 times the angle of projection, which is 45 degrees, that's sin 2 times 45, then over G.

15:57 Then at the end of the day, we have R equal to 4000 meters, 4000 meters.

16:06 Now with the understanding of what we just did, let's look at the questions to know how far

16:15 you can go about a projector or motion.

16:20 Like number one, a body is projected upwards with an initial speed of 400 meters per second

16:27 at an angle of 30 degrees, find about one time of flight to horizontal distance traveled.

16:33 And the second question is, tone is fired from a catapult at an angle of 60 degrees and

16:39 an initial speed of 60 meters per second, calculate the time of flight.

16:47 Number three, what is the maximum range?

16:51 When the angle of projection is, that is, what is the angle of projection for maximum range?

16:59 Number four, what is the projectile? Number five, give two examples of projectile in sports

17:06 and in warfare.

17:08 Thank you for watching, my name is Mr. Bakari, till next time.