00:01
Okay, so this is a projectiles question.
00:02
Now, the only problem is that it says, i suppose it's framed from the same height as in a practice it problem, and i can't see what that height is.
00:10
So i'm going to call that height h, and then you can fill in h and just answer the question yourself from there.
00:17
So the first part asks us for the time of flight.
00:21
So we're going to use, so we have two equations here.
00:24
We have that the horizontal displacement is given by the horizontal velocity times the time because there's no acceleration horizontally, so that's just speed times time.
00:35
And that is u times cos 32.
00:39
That's the horizontal initial velocity times, i'll call it t -end, the time taken to get to the floor.
00:50
And then we have another equation, which is that s -y equals u -y -t, i .e.
00:57
The vertical displacement equals the vertical initial velocity minus a half gt squared because we have gravity accelerating it down.
01:05
And this becomes minus h equals u, sine 32 t -end minus a half gt end squared.
01:18
Now we want to get rid of you because we don't know what you is.
01:21
So we're going to rearrange this first one to find that, because we know sx is 45 .6.
01:29
So we're going to rearrange this first one to find that u is equal to 45 .6 over cos 32 times t -end.
01:41
And if we sub that into this equation down here, we find that we get 45 .6 tan 32.
01:55
And so if we knew h, we could put all this in and solve it.
01:59
But since we don't, i'm just going to have to say that i'm just going to rearrange this and find that t -end is the square root of 2 over g times 45 .6 tan 32 plus h.
02:16
And so when you know h, you can just sub that in and solve it...