00:01
Good day, the topic is about free fall.
00:03
Freefall is a type of motion where the object is being influenced only by gravity.
00:08
As such, it accelerates due to the acceleration, it accelerates at the rate of acceleration due to gravity.
00:16
In this problem, we're given that a stone is dropped into the river that is 43 .9 meters above the water.
00:24
Let's draw a scenario here.
00:25
Let's say this is your bridge, right here, and then below it is the water surface and then the stone is dropped from the bridge that is situated 43 .9 meters above the water surface so this height here is 43 .9.
00:49
Now one second later after the first stone is dropped, the second stone is another stone is dropped.
00:59
Okay, after after t equals one second now the idea here is that the two stones would hit the surface of the water by the same time so that's the about the only possible way for that is the second stone must not just be simply trapped with an initial speed so that it would travel it would catch up with the first stone so we want to find that speed by which the second stone is drop for second stone that's our first problem and for the second one let's try to plot the velocity time graph for the two stones so plot bt graph okay so start by assuming that there is no air resistance so that we assume that the motion is in free fall and the acceleration for both stones is due to gravity so since they would need to travel the same height which is 43 .9, then we can set the two distances traveled by the two stones as equal to each other.
02:24
D1, you see the kinematic equation v1t plus one half a t1 squared will be equated with v2t2 plus one half a t2 squared.
02:42
Since the first stone is simply dropped, it's not given with any initial speed so that v1 there is going to be 0.
02:50
So we have 1 half times a 1 squared equals.
02:54
Now unlike the first stone, there is v2 here.
02:59
So we'll keep at us that times a 2 squared.
03:04
Moreover, we relate t1 and t2 as follows.
03:08
So t2 okay or rather the time of wall of t1 minus the time of wall of t2 is equal to 1 because the two stones are separated at the two stones at the time at which the two stones are dropped is separated by one second so the b1 travels longer amount of time than b or rather stone 1 travels longer amount of times and stone two so here we can write a t -sook as in terms of t -1 as t -1 minus 1 okay so we have one -half times a t -1 squared equals b2 times t1 minus 1 plus one half times a times e 1 minus 1 squared sequence here we have the we have the we have have both v2 and t1 missing well actually b2 is indeed missing but we can solve for t1 so going back to our climatic equation that is d1 equals v1 t1 plus 1 1 1 1 1 1 1 2 well again we have this as the whole t 0 we can solve for t 1 here because we know what d1 is so d1 being for the the height of the a breach is 43 .9.
04:45
So 43 .9 equals one half times the acceleration due to gravity.
04:50
Let's take downward as the positive direction so we keep 9 .8 as possibly.
04:55
So therefore, the 1 is going to be equal to square root of 2 times 43 .9, provided by 9 .8.
05:06
And that produces time 1 as 2 .99...