Parabolic Motion -02- Coordinates of Particles in Projectile Motion

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A particle is projected with a velocity of 30 m/s at 37° with horizontal. Find co-ordinates of the particle after 2 s. Given, g = 10 m/s², sin 37° = 0.6 and cos 37° = 0.8?

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00:00Hi friends, every day of your life is a priceless gift to you from God.

00:06Use every second for self-improvement and humanity.

00:11This question is almost similar to the previous question.

00:15A particle is fired at a certain angle to a flat surface.

00:20We are asked to determine the position of the particle after two seconds.

00:28I think we already know that the particle will perform a parabolic motion.

00:35We don't know where the bullet was fired from, just place that point as the center of coordinates.

00:43The speed of the particle is 30 meters per second, and the elevation angle is 37 degrees.

00:52Let's say after two seconds, the particle reaches point 1.

00:56We can only guess the location of point 1.

01:01Point 1 will have coordinates x1 y1.

01:07Every particle that is around the earth will experience acceleration due to gravity.

01:14We can write the velocity in the form of a unit vector.

01:18The vector v0 can be written as 30 cosine 37 i-hat plus 30 sine 37 j-hat.

01:28The values of sine and cosine are listed on the question sheet.

01:33This is the vector v0.

01:38For vector a, of course, minus 10 j-hat.

01:44We often use vector s as displacement.

01:47Vector s is equal to vector v0 times t plus half of vector a times t squared.

01:56Just substitute the vectors v0 and a that have been obtained previously.

02:03The particle moves two seconds later, meaning t is equal to 2.

02:10Now collect based on the same unit vector.

02:14Vector s is equal to 48 i-hat plus 16 j-hat.

02:21What does this value mean?

02:23The value at i-hat is the abscissa of the particle.

02:26While the value at j-hat is the ordinate of the particle.

02:32So the coordinates of the particle after moving for two seconds are 48 16.

02:38Happy learning everyone!

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