00:01
So we have a child throws a ball, then initial speed of 8 meters per second, 40 degrees above the horizontal.
00:06
The ball leaves a hand, 1 meter above the ground.
00:08
How long is the ball in flights before it hits the ground? so let's first draw a diagram.
00:15
So we have 8 meters per second.
00:19
This is angle of 40 degrees above the horizontal.
00:22
So the y components would be 8, sine 40 degrees.
00:26
X component is 8 costs 40 degrees.
00:30
The height above the ground was one meter and so first one to find how long the ball is in flight before it hits the ground so in the wide direction first going to find the time it takes a rich maximum height so time fruits mountain height we're going to use the equation v is equal to u plus a at maximum height the final velocity is equal to zero that's v here then shall velocity in the wide direction this is the wide direction little bit yeah so u is 8 sine 40 which is done initial velocity the acceleration depends on the acceleration to gravity i'm taking up as positive so down to be negative so this is if you're trying to gravity is accident downwards that's minus 9 .8 times time t so on for t there this will be called to 0 .5247 seconds so the time to reach the maximum height and then come back down to the same heights was launched from be two times stat so this will be 1 .0494 seconds so now we need to find again the time it takes the 4 from this height to ground and when the ball gets back down to this height is going to have the same velocity it had when it was going up initially it is 8 sine 40 in the opposite direction downwards so we're going to use s is equal to u t plus half a t squared and this is also in the wide direction so we have s as displacement which is a downward displacement so taking down as negative that's minus 1 is equal to u is going downwards that it minus 8 sine 40 times the time to go from down there to that's the time one to find plus half exertion is due g which is minus 9 .8 t squared so we're solving for t here so we get two times here we get a positive and then negative value so we want to use a positive value because time can be negative so this 0 .149 4 seconds so adding that to our original time get the total time scroll this t total is equal to 1 .0494 plus 0 .14 plus 0 .14 94 which is your code say 1 .198 8 seconds approximate to 1 .199 1 .1 .1 .1 2nd 2 seconds sir so for b how far from where the child is standing does all hit the ground so that'll be the horizontal range and in the x direction as well we'll go to next we have a constant velocity lost t in the extraction is equal to disillismance the excursion divided by the time just the biggest t total found here so i've lost in the extractions is called it costs 40 is equal to d divided by t which is 1 .1988 so this is equal to d will be water that is 7 .35 met as approximately i'm sorry it wasn't a calculation here this should be 0 .167 seconds so our total here is equal to 1 .271 to approximately 1 .22 so this is 1 .271 so this is 1 .271 which should be 7 .46 meters instead for the displacement now for c, what's the max height that the ball reaches? so, from the point where it was launched, once you use the equation, s, which are v squared is equal to u squared plus 2 as.
05:41
As is a pier out displacement in the wide direction.
05:44
This is the way direction...