00:01
Here we have to solve the following problem.
00:05
In equation a, we have to determine the launch speed.
00:09
Let's call it v0.
00:12
So here, the angle of the launch is 45 degree and actually this initial velocity is called ve.
00:26
Let's follow this notation.
00:28
So the height of the gate is 0 .81 meter and the gate is at the horizontal distance.
00:37
Of x which is 1 .5 meters.
00:40
It means that here if we show the flight we can notice the following.
00:55
Here x equals to v0 cosine alpha times t and therefore we can express this time.
01:12
On the other hand height h is the initial velocity times alpha times time minus gt squared over 2.
01:28
Now we can determine this time, let's do this.
02:23
And now we can determine ve.
02:52
That is 5 .65 meters per second, which is roughly 5 .7 meters per second.
02:59
Now we can move on to question b...