A rocket is launched at an angle of 53.0^∘ above the horizontal with an initial speed of 100 m

A rocket is launched at an angle of 53.0^∘ above the...

Key Concepts

Gravity

Gravity is the force that attracts objects toward the center of the Earth, providing a constant downward acceleration that acts on the object during its flight. In projectile motion analysis, gravity is the only acceleration acting on the object once external propulsion ceases, affecting the time of flight, maximum altitude, and horizontal range.

Acceleration Phase

The acceleration phase refers to a period during which the object moves under a specific, non-gravitational acceleration, such as when a rocket’s engines are active. This phase is analyzed separately from the projectile (free-fall) phase, since the initial conditions at the end of the acceleration phase serve as the starting point for the subsequent projectile motion.

Projectile Motion

Projectile motion refers to the motion of an object that is thrown or projected into the air, subject only to the acceleration of gravity. This concept involves analyzing the vertical and horizontal components of motion separately, as gravity affects only the vertical motion while the horizontal motion remains uniform (in the absence of air resistance).

Vector Decomposition

Vector decomposition is the process of breaking a given vector into its orthogonal components, typically along the horizontal and vertical axes. In the context of projectile motion, the initial velocity is decomposed into horizontal and vertical parts using trigonometric functions, which then allows separate analysis of the object's motion in each direction.

Kinematics

Kinematics involves using a set of equations that describe the motion of objects under constant acceleration. These equations are crucial for calculating variables such as displacement, velocity, and time when the acceleration is known, as in the powered and projectile phases of the motion.

Transcript

00:01 So this question involves a situation in which there is a constant acceleration motion in one direction followed by a projectile motion.

00:16 So what we'll do is we'll split up the motion into those two parts and we can answer the questions about the entire motion using that.

00:26 So what we have then is something that is moving in a constant direction, 53 degrees above the horizontal, and then after a certain amount of time, i believe three seconds, then the object will lose its engine push and so then it just becomes a projectile.

00:49 It actually becomes a projectile.

00:51 It's going to eventually fall back down to the ground again and so there's going to be three separate parts of the motion that we're going to need to look at.

01:04 There is this part of the motion when the object is going up in a straight line.

01:10 We need to determine this distance that the object goes up and the distance the object goes forward by finding this distance here and then using the angle of 53 degrees to find out what those two components of that motion are.

01:25 Once we get that motion, we can find out how high the object rises above that y naught starting position and call that delta y and then y naught plus delta y be the total height that the object reaches and then the object is going to fall back down so we can find out with delta y plus y naught again we can find out how long it's going to take to reach the ground and from that we can find out the total motion in the horizontal direction which is going to include this little bit right here and then this little bit and then the original x naught.

02:11 So that's our goal and we'll start off by trying to find what d is.

02:17 D, the distance it travels in a straight line, would be given to us by a constant acceleration equation.

02:24 We can say that that distance is going to equal the initial velocity which we're told is 100 meters per second i'll leave off the units for clarity times the time of three seconds plus one half times the acceleration which is 30 meters per second squared times the time squared which is three seconds squared so this is 300 meters plus 30 times a half is 15 times three squared is nine so that's 15 times nine that's 135 so this is 435 is the initial distance d that the rocket rises up into the air.

03:17 Now we need to get the components then x naught and y naught.

03:21 Y naught is going to the y component of that motion so it's going to be 435 meters times the sine of 53 degrees.

03:34 So you put in a calculator 435 times sine of 53 and you get that y naught is equal to 347 .4 meters.

03:49 X naught is going to be the x component of that motion that's going to be 435 times the cosine of 53 degrees.

04:02 Move this out of the way.

04:04 And so that ends up being 435 times cosine 53 that gives us 261 .79 meters.

04:20 Now we also need to know the speed that the projectile has when it begins its projectile motion so we're going to need to use an equation for the one -dimensional motion of the rocket as it's rising with the motors.

04:37 The velocity it'll have at the end of that motion is going to be the velocity it has at the beginning of the motion which was 100 meters per second plus the acceleration time to time the acceleration was 30 meters per second and the time was three seconds.

04:56 So the velocity you'll have at the end of that motion is going to be 100 plus 90 so it's going to be 190 meters per second.

05:05 Again we need to find the components of that velocity so we will say that the y naught the initial y velocity in when the object becomes a projectile is going to equal 190 times the sine of 53 degrees and it's like 190 times sine 53 that comes out to be 151 .7 meters per second and then the x velocity after it becomes a projectile is going to be 190 times the cosine of 53 degrees.

05:53 You put that in your calculator and that works out to be 190 times cosine 53 degrees that works out to be 114 .344 meters per second.

06:07 Now we're ready to start the projectile motion part.

06:11 Let's solve for this delta y distance here first.

06:15 Actually let's solve for the time it takes to reach that distance that's one of the ways to find that distance.

06:21 We know that at the when it's at its maximum height the vertical velocity is going to be zero so we can use this equation that says v is equal to v naught plus at we can use that equation but we can say the velocity when it reaches maximum height is going to be zero and this is going to be the initial velocity in the y direction and the acceleration is a negative g so this is a negative g times t so that gives us that v naught y over g is equal to the time it takes to reach that maximum height after it becomes a projectile...

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