0:00
So here we have some givens.
00:01
We have an angle of launch of 45 degrees.
00:05
We have a delta y of zero meters.
00:09
So essentially where the jumper is landing is the same horizontal plane as the plane in which she is jumping.
00:23
So therefore there is no height difference in her jump to her landing.
00:29
We have a delta x of 8 meters, and we can start off by using the range equation.
00:40
Because delta y equals 0 meters, we can use the range equation in order to find out the initial velocity.
00:49
So r equals range, which essentially equals delta x, and this will equal the initial squared sine of 2 theta.
01:01
Over g now sign of two theta theta is 45 degrees two times 45 degrees is 90 and we know that sign of 90 degrees is going to equal one so we can essentially say that the rather that the initial velocity is going to be equal to the grab the acceleration to the graphy times delta x to the one half power so the initial will equal 9 .81 times 8 to the 1 half.
01:39
And this is going to equal 8 .85 meters per second.
01:45
So this would be our essentially our initial velocity at launch.
01:51
And then for part b, they're asking us, delta x now equals 10 meters...