00:01
Hello, a baseball passes a 3 meter tall wall, which is at the distance of 110 -20 meters from the home plate.
00:17
The baseball leaves a bed, making an angle of 35 degree with respect to the horizontal, and at the initial height of 0 .80 meters above the ground.
00:30
First, we have to find the initial speed.
00:34
Let's solve this problem.
00:36
Here we will sketch this short in y and x -axis.
00:40
So let's label h0.
00:43
And we will label also the initial angle alpha.
00:50
And now it's time to show the trajectory.
00:53
The trajectory is parabolic.
00:56
And the wall, which is cleared, is located at the distance d and at the height, h, above the ground.
01:06
Let's write down the equation for why is the function of time.
01:09
That is h0 plus v0 sine alpha t minus gt squared over 2 and x as a function of time is v0 sine alpha when t equals to t1 x becomes d therefore d equals to v0 sine alpha and thereby t1 is d over v0, sine alpha.
01:58
Let's now simplify, let's now make a substitution in equation 1.
02:05
Here, y at the moment of t1 equals to h.
02:11
And therefore, that is h0 plus v0.
02:20
Oh, i think i've made the typo.
02:22
Yeah, of course.
02:24
In d, yeah, here, x coordinate, of course, must be multiplied by cosine of alpha.
02:32
So let's fix it because the horizontal component of the speed of the velocity to be precise vx is v0 cosine alpha.
02:45
So let's fix this typo and now let's complete the calculation...