A projectile is launched with an initial speed of 30 m / s at an angle of 60^∘ above the horizon

A projectile is launched with an initial speed of 30 m / s...

A projectile is launched with an initial speed of 30 $\mathrm{m} / \mathrm{s}$ at an angle of $60^{\circ}$ above the horizontal. What are the (a) magnitude and (b) angle of its velocity 2.0 s after launch, and (c) is the angle above or below the horizontal? What are the (d) magnitude and (e) angle of its velocity 5.0 s after launch, and (f) is the angle above or below the horizontal?

A projectile is launched with an initial speed of 30 $\mathrm{m} / \mathrm{s}$ at an angle of $60^{\circ}$ above the horizontal. What are the (a) magnitude and
(b) angle of its velocity 2.0 s after launch, and (c) is the angle above
or below the horizontal? What are the (d) magnitude and (e) angle of its velocity 5.0 s after launch, and (f) is the angle above or below
the horizontal?

Key Concepts

Projectile Motion

Projectile motion describes the trajectory of an object that is launched into the air and moves subject to only the acceleration due to gravity. This motion is characterized by a parabolic path, where the horizontal and vertical components of motion are treated independently, with the horizontal motion having constant velocity and the vertical motion having uniform acceleration.

Vector Decomposition

Vector decomposition involves breaking down a vector into its horizontal and vertical components using trigonometric functions. In the context of projectile motion, the initial velocity is split into these components based on the launch angle, allowing the analysis of motion separately in the horizontal and vertical directions.

Acceleration due to Gravity

Acceleration due to gravity is the constant acceleration that acts on an object in free fall near the Earth's surface, typically taken as 9.8 m/s² downward. This acceleration influences the vertical component of the motion, causing the projectile's vertical speed to decrease as it rises and increase as it falls.

Kinematic Equations for Uniform Acceleration

Kinematic equations for uniform acceleration relate the displacement, velocity, acceleration, and time for objects moving with constant acceleration. These equations are crucial in projectile motion to determine the vertical velocity at any given time, while the horizontal velocity remains unaffected by gravity.

Resultant Velocity Calculation

Resultant velocity calculation involves combining the horizontal and vertical components of velocity to determine the overall speed and direction of the projectile. The magnitude is obtained using the Pythagorean theorem and the angle can be calculated using an inverse trigonometric function, indicating the direction of the motion relative to the horizontal.

Transcript

00:01 Part a, in unit vector notation, the first displacement delta r sub 1, this is going to be equaling to 60 .0 kilometers per hour.

00:13 We are going to multiply this by 40 .0 minutes divided by 60 minutes for every hour.

00:27 I -hat and so this is giving us 40 .0 kilometers i -hat and so the second displacement we can say r -sup 2 or other delta r -sip 2 this is going to be equaling to 60 .0 kilometers per hour multiplied by 20 minutes divided by 60 minutes for every hour and this is giving us 20.

01:04 Kilometers and it's more appropriate to say the magnitude of the second displacement because its direction is 40 degrees north of east and so with we can say delta r sub 2 in unit vector notation would be 20 .0 kilometers multiplied by cosine of 40 degrees i hat so that be our x component plus 20 0 .0 kilometers multiplied by sign of 40 degrees j hat.

01:46 And so we find that the second displacement in unit vector form is going to be 15 .3 kilometers ihat plus 12 .9 kilometers j hat.

02:02 The third displacement then is equaling negative 60 .0 kilometers per hour multiplied by 50 minutes divided by 60 minutes per hour and then this would be i -hat and this would give us negative 50 .0 kilometers i -hat and so the total displacement we can simply say delta r would be the sum of all the the first second and third displacements and so this would be giving us 40 .0 kilometers i -hat plus 15 .3 kilometers i -hat plus 12 .9 kilometers j -hat minus 50 .0 kilometers i -hat.

03:18 And so we can say that then our total displacement in unit vector notation would be giving us 5 .30 kilometers i -hat.

03:30 Plus 12 .9 kilometers j hat.

03:37 The total time for the trip, delta t, would be simply 40 minutes plus 20 minutes plus 50 minutes, so 110 minutes.

03:50 And we can then say that this is approximately equivalent to 1 .83 hours.

03:56 And so we can then find for part a, the average velocity vector, this would be equaling to 5 .3 hours...

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