A particle is projected with a velocity of 20 m/s at 53° with horizontal. Find velocity of the particle after 2 s. Given, g = 10 m/s², sin 53° = 0.8 and cos 53° = 0.6.
watch other videos related to PARABOLIC MOTION:
https://dailymotion.com/playlist/x8xg5w
#[arabolicmotion
watch other videos related to PARABOLIC MOTION:
https://dailymotion.com/playlist/x8xg5w
#[arabolicmotion
Category
📚
LearningTranscript
00:00Hey guys, unless you start looking for inspiration within yourself, you will not achieve long-term
00:06inspiration to pursue monumental things in life.
00:12The question on the screen is the first in the projectile motion series.
00:17A particle is fired with an initial velocity of 30 meters per second at a certain angle.
00:23Now, we are asked to calculate what the velocity of the particle is after moving 2 seconds
00:29later.
00:33Let's discuss this question in more detail.
00:37A particle fired at a certain elevation angle to the ground surface will go through a parabolic
00:42path.
00:46In general, we assume the starting point of the particle is at the center of coordinates.
00:54At this point, the particle has a velocity of V0, and the elevation angle is 53 degrees.
01:04Let's say after 2 seconds, the particle is at point 1.
01:07The particle's velocity changes to V1.
01:13All particles around the Earth's surface will experience acceleration due to gravity.
01:21Velocity and acceleration are vector quantities.
01:24The vector of the initial velocity is V0 cos 53 i-hat plus V0 sin 53 j-hat.
01:35We can see the value of V0 from the question sheet.
01:41The vector V0 is 12 i-hat plus 16 j-hat.
01:48The vector of the particle's acceleration is minus g j-hat.
01:54The magnitude of the acceleration due to gravity is 10 meters per second squared.
02:02These two quantities are related by the equation, vector V1 is equal to vector V0 plus vector
02:08A, times T.
02:11This is the general equation for the motion of a particle with constant acceleration.
02:19We can substitute vector V0 and vector A into this equation.
02:26We are asked to find the value of the particle's velocity when the particle has been moving
02:30for 2 seconds.
02:35Vector V1 is 12 i-hat minus 4 j-hat.
02:41The magnitude of the velocity at point 1 is the norm of vector V1.
02:48It is about square root 160, or 4 square root 10 meters per second.
02:55This is the answer to this question.