A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 100 m above the glacier at a speed of 150 m/s. How far short of the target should it drop the package?

Video transcript

Hey everyone. Welcome back in this problem. A pickup truck moving with a speed of 80 m per second is passing over a bridge situated at a height of 50 m above water below the bridge is a river in which a person is sitting in a steady boat, person in the trucks, cargo bed stretches their hand out off the bridge to deliver a parcel in the boat. And we're asked to determine the horizontal distance from the boat where the person in the truck should release the parcel so that it falls exactly in the boat. We're given four answer choices. Option A 319 option B 256 option C 100 option D 50. And those are all in meters looking for a horizontal distance. We're told some information about speed. So we want to think about our kinematic or U AM equations here. So let's go ahead and write out the variables that we have starting in the X direction. Now, the person in the truck's cargo bed is gonna reach out and release the parcel. Ok? So when they release that parcel have an initial speed affected by the speed of the truck. Ok. So they're not throwing it with any initial speed, they're just releasing it. It's gonna have the same initial speed as this track. And so in the horizontal direction, the initial speed is going to be m per second, same as the truck. Now, the horizontal distance, we don't know this is what we're looking and we don't know how long. Now in the X direction, we're gonna assume that there's no acceleration and there's no drag or anything like that. So we only have these three very let's now move to the Y direction. OK? Because in the X direction, we only have one known value. We can't use that to find delta in the right direction. The initial velocity of V not Y is going to actually be m per second. This person is going to release, drop this parcel. So the initial vertical speed is zero. OK? The truck has no vertical speed, the person's not throwing it. So the initial vertical speed is zero, we don't know the final vertical speed know that the vertical distance traveled delta Y is going to be m and double to why the displacement is actually going to be negative and we're gonna take upwards to be positive. So if we're on top of the bridge and that parcel falls down, it's going in the opposite direction, the negative direction. And so that's gonna be a displacement of negative 50 m the acceleration is gonna be the acceleration due to gravity which is negative 9.8 m per second squared. And we don't know the time to now what we want to find is delta X. In order to find delta X, we need the time T. Now the time T that it takes in the extraction is the same as in the wider the horizontal and vertical times are going to be exactly the same. So if we use the vertical information to calculate the time T, then we can use that in our horizontal equation to find delta. So starting in the Y direction, we have three known values V naught delta Y and A, we're looking for the time T, so we choose the equation that has those four variables and that's gonna be delta Y is equal to V naught Y multiplied by T plus one half multiplied by a multiplied by T squared. Substituting in our values, we have negative 50 m is equal to one half multiplied by negative 9.8 m per second squared, multiplied by T squared, dividing both sides by one half multiplied by negative 9.8 m per second squared. We get that T squared is equal to 10. 2nd squared. Approximately taking the square root. We get that the time T is approximately equal to 3.1944. I know when we take the square root we get both the positive and the negative route, but we're looking for time. So all we care about is that positive route. And so we're gonna take that positive answer. OK. So we have our time tea here. Now we need to go ahead and use that in our horizontal equation to find the horizontal distance. Now, in the horizontal direction, we have that the velocity V no X which is just consistent throughout the motion. It's gonna be equal to delta X divided by TK velocity equals distance over time. That means that this horizontal distance delta X is gonna be V not X multiplied by T. So substituting in our values, delta X is gonna be equal to 80 m per second multiplied by 3.1944 seconds which gives us a horizontal distance of about 255. m. Now rounding this the horizontal distance from the boat where the person should release that parcel is gonna be 256 m which corresponds with answer choice. Thanks everyone for watching. I hope this video helped you in the next one.