00:02
Here we have an archer who fires an arrow with a given initial velocity and launch angle and we have to find for part a the horizontal component of initial velocity.
00:15
So initial velocity component is v0 v0 x equal to v0 cos theta.
00:21
V0 is initial horizontal velocity 42 meter per second.
00:27
Cos theta is the angle of launch which is 35 degree.
00:30
So this gives us 34 .4 meter per second.
00:40
Now for part b we have to find out the vertical component of initial velocity.
00:46
So it is v -o -y equal to v0 -synt -sin theta equal to 42 meter per second, sine 35 degree.
00:56
So this gives us 24 .1 meter per second.
01:09
For part c we have to find out the maximum height attained by the arrow so using equation of kinematics let us say at maximum height the vertical velocity is vfy so vfy square equal to v0y square plus 2 a y s where s is the displacement from initial to maximum height so final velocity in vertical direction at maximum height will be zero initial velocity is 24 .1 meter per second in vertical direction plus 2 acceleration is acceleration due to gravity so it will be minus 9 .8 meter per second square because positive y x is upward so acceleration will be negative because it is downward and s is the displacement from initial point to the maximum height so in place of s we can right h, h is the maximum height, so this gives us h equal to 29 .6 meters.
02:29
For part d we have to find out the time taken to reach the maximum height.
02:37
So we again use equation of kinematics for vertical motion.
02:42
Vfy equal to v0y plus ayt.
02:46
At maximum height, vertical velocity is 0.
02:52
V0y the initial vertical velocity plus acceleration in y direction minus g into t.
02:58
So t gives us v0y point g equal to 24 .1 meter per second divided by 9 .8 meter per second square...