00:01
Here for the solution, use conservation of energy to find the velocity just before leaving the track.
00:09
That is half mv i .2i2 plus m ghe i equal to half mvf square plus m gh f.
00:17
As sky jumper starts from rest, also cancel m on each side.
00:25
So we get 0 plus 9 .8 multiply by 40 .5 is equal to half v square plus 9 .8 multiply by 19 .5.
00:37
So from here we get vf equal to sqrt in bracket 2 multiply by 9 .8 multiply by 40 .5 minus 19 .5.
00:47
And from here we get vf equal to 20 .287 meter per second.
00:53
Now to find the maximum height attend, we must understand the fact that the maximum height height speed is zero momentarily.
01:09
So using energy conversion, we have half mv square equal to m zh.
01:15
Now note that the launch angle is 45 degree.
01:21
So we have half multiply by 20 .287 sine 45 whole square equal to 9 .8 multiplied by h and from here we get h equal to 10 .5 meter.
01:33
So the total height from the ground, that is s total, will be equal to 10 .5 plus 19 .5 equal to 30 meter.
01:44
That is maximum height attained above the ground...