[Music] welcome back let's do another example this time where we have constant acceleration this question is a little bit more complex let's give it a go so the question a sandbag is dropped from a balloon that is ascending vertically at a constant speed of 6 m/s the bag is released with the same upward velocity of 6 me 6 m/s at T is equal to 0 seconds and hits the ground when T is equal to 8 seconds find the speed of the bag as it hits the ground and the altitude of the balloon at this instant so there's a lot going on with this problem so maybe it'll be a little bit easier if we start by drawing like a little quick picture of what's happening so let's say we have our balloon let's draw our balloon and it is traveling upwards at 6 m/s and this is at we'll call time T is equal to0 and after some time later our balloon has moved upwards and we'll call the position s is equal to 0 from the initial position and we'll say the position upwards is s of the balloon and right at the beginning instant the balloon dropped the sandbag and and that dropped down to the ground and it took 8 seconds to get to the ground so let's try and draw our sandbag here so our sandbag starts with the balloon at T is equal to zero and then right at that instant it's dropped and it lands on the ground sometime later when T is equal to 8 Seconds and let's call our position to where that sandbag is on the ground s of the bag so there's one thing that's missing from here and that is uh the acceleration so if this is we're assuming this is on Earth and on Earth acceleration would be downwards and we'll call this G and that's equal to 9 81 squar so we have all of our information uh at time T equals Zer the bag and the balloon are together moving upwards at 6 meters per second and at that instant we're going to say that that is when s our position is equal to zero and then at some time 8 seconds later uh we have the balloon has traveled upwards and the sandbag has landed on the ground and we need to find the position of the balloon the position of the bag and we're also going to try and find the speed of the bag as it hits the ground so we're ask for the speed of the bag as it hits the ground which is the velocity of the bag and we're ask for the altitude of the balloon so the altitude of the balloon is going to be the addition s balloon plus s bag that's the distance of the balloon from the ground so how can we begin well if we use our uh equations of motion for constant acceleration we can solve for the position of the bag and the position of the balloon after 8 seconds has gone by so let's start with the bag so from our equations we can write the position of the bag is equal to s0 which is just zero plus the initial velocity of the bag time t plus 12 times our constant acceleration term time t^2 let's make sure that that looks like a t so we solve it substitute in the values that we know we know the velocity of the bag is the same as the balloon when it's released and that's upwards 6 m/s so let's assume that the upwards direction is going to be positive so we know zero is our s0 our initial velocity is 6 m/s the amount of time that goes by is 8 seconds plus2 our acceleration is 9.81 m/s squar and that's acting downward so that's negative 9.81 and again our time is 8 seconds and if we solve for this the position of the bag after 8 seconds when it hit to the ground we have moved 2 65.9 m in the negative Direction so we found the position of the bag now we can find the position of the balloon if we use the exact same equation but the balloon is going at a constant velocity our acceleration term is just equal to zero so we have S of our balloon is just equal to s0 which is 0 plus the velocity of our balloon which is six meters per second times 8 seconds and then our last term is just zero because the acceleration is zero so we're left with the S of our balloon is 6 * 8 which is 48 met and we can write up and for the bag we can write down so the total position the total altitude is equal to S of the balloon position of the balloon minus the position of the bag which is equal to 48 minus 265.0 which is approximately equal to 314 M so when the bag hits the ground the balloon is 34 m above the ground so just remembering that the altitude is the distance from the ground to the balloon and it's not s of the balloon or the position of the balloon because we decided earlier that the position of the balloon at s equal Z is going to be when uh the balloon releases the sandbag so the last question we need to know is find the speed of that sandbag so we can use our other equation of motion for constant acceleration to find the speed of the bag speed of the bag and we'll call this V of the bag and that's equal to V 0 plus acceleration term times time so we know v0 was six m per second and our acceleration is 9.81 our time is 8 seconds and if we solve this we get the velocity is equal to 72.5 m/s and just to keep things simple we can highlight all of our Solutions so our three answers are the altitude sorry two answers the altitude is 314 m of the balloon and the bag hits the ground at a speed of 72.5 m/ second [Music]