0:00
High.
00:01
Here in this given problem, a projectile is thrown horizontally from a height with the initial speed and that initial speed is horizontal means we can say along x -axis.
00:21
So that is given as v i x is equal to 15 meter per second and we know as there is no force, no acceleration along x -axis.
00:33
So it will remain same throughout the motion and its initial vertical velocity that is zero time given is 2 .5 second after which we have to find its final velocity along with the direction so magnitude of vertical velocity vertical component of the velocity for which we will consider vertical motion of this projectile.
01:27
So taking vertical motion and using first equation of motion here, vfy, final vertical velocity that will be equal to v, i, y, plus g because acceleration here is acceleration due gravity g into t means this is 0 plus 9 .8 into 2 .5 means it comes out to be equal to 24 .5 meter per second and already we have seen that final horizontal velocity will be equal to initial horizontal velocity means this is 15 meter per second so so magnitude of the velocity after 2 .5 seconds is v is equal to using pythagoras theorem, that is square of vfx plus square of v fy and having a square root of it.
02:54
So this is 15 square plus 24 .5 square.
03:00
Means this is 225 plus 600 .25 means this is 825 .25...