00:01
Here in this given problem this is the point from which the ski jumper takes a jump.
00:13
This is angle alpha below the horizontal at which he jumps and this alpha is given as 9 .5 degree and his initial velocity v0 that is 28 m per second.
00:29
And then this is the slope, the inclined slope and he jumps such that at the ground his final velocity vector it should be making an angle beta with the ground and that beta is just 3 .0 degree.
00:55
So we have to find the angle let it be gamma which the ground should be making, which this slope should be making with the horizontal.
01:05
So for which we will find the total angle theta, total angle made by the final velocity with the horizontal surface this.
01:16
In order to get that we should have horizontal component of final velocity and vertical component of final velocity provided horizontal initial velocity that is vox and vertical velocity initial velocity voy.
01:36
This horizontal distance covered that is given to be 55 meter.
01:44
So first of all using the concept that horizontal motion in a projectile motion remains uniform so vfx that will be equal to vox means vocos alpha means 9 .5 degree.
02:01
V0 that is 28, 28 multiplied by cos 9 .5 degree and it is calculated to be equal to 27 .6 meter per second and that will remain uniform throughout the motion...