00:01
A burglar is trying to evade the police by jumping off from the top of a building to another rooftop of a building that is three meters below.
00:12
He can cover the 50 meter dash in 5 .6 seconds and we'll take that as the maximum speed and also the initial speed of the person here as a projectile.
00:27
If he jumps horizontally initially, then the y component of his initial velocity is zero.
00:36
Whatever is the x component, that is also the initial velocity, and we take it as his full speed.
00:42
So the question now is, can he make it across at full speed if the distance in the alley is 8 meters? so we need to determine first the time of travel of the burglar here.
00:57
And within that time interval we will solve for the horizontal distance.
01:02
If the horizontal distance xf2 is equal or greater than the alley distance 8 meters, then he will make it across.
01:10
Otherwise, he falls down.
01:12
So let's calculate for the time of travel first of the robber here.
01:20
Okay, since he is considered as a projectile, we neglect air resistance.
01:25
Then as a projectile, his acceleration along the y -axis is due to gravity, which is constant.
01:33
So we choose among the constant acceleration along the y -axis, and the most practical one would be this equation.
01:41
We have the final position along the y is equal to initial position along the y.
01:47
Let's label the initial as subscript 1, plus v1y -t, minus, minus, minus, minus one half acceleration due to gravity times t squared.
02:00
Let this be zero.
02:02
And the final y here, since we have the reference here at the rooftop of the lower building, then the final height of the burglar here will be considered zero and the initial is three.
02:18
So we'll take this as zero.
02:20
Then algebraically, we transpose y sub one to the left hand side.
02:25
It becomes.
02:25
Negative y sub 1 and then we still have here negative one half g t so we multiply this by 2 get reading of that and then we multiply divide this by negative g so that this now becomes okay this is now t squared okay we're not changing the sign of negative 1 half g here and then we get the square and then we get the square root of course of both sides then we put in the values now two times three meters negative negative makes it positive and then we have 9 .8 here therefore it will take 0 .78 seconds okay if we have by the way why is v1y 0 here because the initial velocity of the person here has no y component and and if we, the full speed here is just distance over time.
03:33
So we have 50 meters covered in 5 .6 seconds.
03:38
So his full speed is actually 8 .9 meters per second going east.
03:44
Okay.
03:46
That's why.
03:47
So in this time interval, let's determine his horizontal distance covered.
03:56
So we will just be using this equation.
04:02
For a projectile, there's no acceleration in the x -axis.
04:06
So we have your final distance along the x equals initial plus v1x times time, plus one -half a xt squared, which is of course 0.
04:20
And we take this also a 0.
04:22
So we have v1x, as mentioned earlier, is 8 .9.
04:26
And then we have the time of travel which is 0 .78 so he is covering 6 .9 meters obviously at 0 .78 he is already he has already covered the 3 meter distance along the y but the x is just 6 so therefore definitely he's not going to make it because he's just somewhere here where the distance here is just 6 .9 meters.
05:06
So the answer therefore, no, he's not going to make it...