00:01
Hello guys, so for the given problem we have been provided that a projectile is launched horizontally from a top of a building having height h is equal to 30 meter.
00:11
The velocity of projection of the projectile u is given as 30 meter per second.
00:17
The acceleration due to gravity g is 10 meter per second square acting downwards.
00:23
Here we need to find the final velocity of the projectile just before hitting the ground, which we are denoting with the letter b.
00:31
Okay, so here we know that the initial velocity vector that is u vector which is equal to ux i cap plus uy j okay as the initial velocity in y direction is zero so u vector is equal to ux i cap having magnitude having magnitude equal to 30 meters per second.
01:16
All right? and the final velocity vector, that is v vector, will be given by vx i cap plus vyj cap.
01:34
As the acceleration in x direction is equal to zero, hence vx will be equal to ux which is having the value 30 meters per second but here we do not know the value of the y component of the final velocity here so to find it first of all we need to determine that time of flight so the time of flight t is given by the formula under root 2h divided by g where h is the height of the building and g is the acceleration due to gravity...