00:01
Hello, here we have to solve the following problem.
00:02
Ball leave the butt with a speed of 33 meters per second at an angle of 37 .3 degree above horizontal.
00:11
First in question a, we have to determine at which two times the height is above 10 .2 meters.
00:22
Here, let's make a quick sketch.
00:32
This motion occurs only under the gravity, that's why.
00:35
Why is v0.
00:37
Sign alpha t minus gt squared over 2 and that means that 4 .90 squared minus 20t equals to 10 .2.
01:07
Let's find the discriminant and now let's calculate times.
01:26
Let's calculate t 1 and 2.
02:10
And now let's move on to question b.
02:17
We have to return it horizontal component of the velocity vx.
02:22
That's v0 cosine theta.
02:46
Now in equation c, we have to calculate vertical component of velocity at both times.
03:00
And here, vy is a function of time is v0, sine alpha t minus gt.
03:12
B0 sign out for minus gt...