00:01
Hi there, here we have to find the initial take of speed of an insect which leaps and reaches the maximum height of 58 .7 centimeters during the jump.
00:15
And also we have to find the travel horizontal distance.
00:19
So let's illustrate the situation.
00:21
Let's introduce y and x x x.
00:25
And the origin is the initial point in the point where the insect jumps.
00:32
So the insect jumps at the angle of 58 degree above the horizontal.
00:39
That is roughly here and that's the initial velocity.
00:46
The insects abase parabolic trajectory and reaches maximum point y maximum somewhere in the middle of the travel.
01:05
And then lands and we have to also calculate this displacement at this distance l and initial velocity v0 let's do this so first of all let's find projections of v0 on x -axis and y -axis v0 x x equals to v0 times cosine alpha and v0 y sine alpha.
01:43
There is only one force which exists in the system that is the gravity.
01:50
And that's why vx is constant and that equals to v0 x, which is v0 times cosine alpha.
02:00
Meanwhile, vy depends on speed, depends on time, and there is v0y minus gt, which is v0, v0, which is v0, sine alpha.
02:14
For minus gt now let's see when this maximum flight is when this point is reached this point is reached when vertical component of the velocity y is zero so therefore at some moment of moment of half of the flight, c1 half, vertical component is of speed is 0, and that equals to v0 sine alpha minus g, c 1 half.
03:17
Therefore, c 1 half equals to v0 sine alpha, divided by g...