Parabolic Motion -03- Perpendicular Velocity In Parabolic Motion

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A particle is projected with velocity 30 m/s at angle 30° with horizontal. Find the time when the velocity vector is perpendicular to the initial velocity vector.

Transcript

00:00Hi guys, good morning. I hope you have a good day ahead.

00:07By the way, in front of your screen is a pretty interesting problem.

00:12A particle is launched with an initial velocity of 30 meters per second at an angle of 30

00:17degrees to the horizontal. Well, we are asked to estimate the time when the particle's

00:24velocity is perpendicular to its original velocity. Let's analyze this case.

00:34First of all, we will create a parabolic trajectory on the Cartesian plane, where the particle's

00:40launch point is right at the center of coordinates. At point 0, the particle's velocity is V0.

00:52The angle of the initial velocity is beta 0. We will extend this velocity line to infinity.

01:02Next, we will draw a line perpendicular to that line. Slide the perpendicular line until

01:11it touches the parabolic trajectory. Let's call that point 1.

01:19The particle's velocity at point 1 will be perpendicular to the particle's initial velocity.

01:27We don't know where this point is located. We are only asked to get when the particle

01:31reaches point 1. What comes to our mind? That's right. Dot product. The dot product of two

01:41perpendicular vectors is 0. Just right. Vector V0 dot vector V1 is equal to 0. Based on the

01:53kinematics of motion, vector V1 is equal to vector V0 plus vector A, t. This equation

02:03can be written like this. Vector V0 dot vector V0 is V0 squared cosine 0 degrees. Because

02:14the same vector is a vector in the same direction. The value of cosine 0 degrees is 1. A here

02:27is the acceleration vector. The direction of vector A is towards the negative y-axis.

02:36Note the angle formed is vector A and vector V0. The angle is 90 plus beta 0. This is correct.

02:46Thus the result of the dot product is V0GT cosine of 90 plus beta 0. Here we do not give

02:55a negative sign to the acceleration due to gravity because the sign is already represented

03:00by the angle between the two vectors. From the trigonometric identity table, the cosine

03:07of 90 plus beta 0 is equal to the minus sign of beta 0. V0 squared minus V0GT sine beta

03:160 is equal to 0. From here, t is equal to V0 over G sine beta 0. Both of these values

03:29are written on the problem sheet. Everyone knows that the sine of 30 degrees is half.

03:40T is equal to 6 seconds. This is the time required for the particle to reach the point

03:47where the velocity at that point will be perpendicular to the initial velocity.

03:54Happy learning everyone!

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