A rifle is aimed horizontally at a target 50 m away. The bullet hits the target 2.0 cm below the aim point. (b) What was the bullet's speed as it left the barrel?

Video transcript

Everyone in this problem during a film shoot, a stuntman has to jump off a moving car which takes off from the cliff and crashes on the ground 30 m away. After 1.5 seconds. Now, we're gonna come back to what this question is asking us to find. But first, let's draw out what we've been given so far. So we have a cliff and we're assuming this is a horizontal cliff and we're not told anything about a ramp or anything like that. We have a car driving on it. We're gonna assume this horizontal cliff. We have a stunt man in this car which we're gonna just draw as a circle, he's gonna drive off of this cliff. And because this cliff is horizontal, that initial takeoff speed of the car, which we're gonna call Vnat X is gonna be completely in the horizontal direction. It's gonna continue downwards until it reaches the ground. And we're told that it reaches the ground 30 m away from the cliff at a time t of 1.5 seconds later. All right. Now, we're asked to determine the speed of the car as it takes off from the cliff can assume that there is no air resistance. Well, the take off speed for this car off this cliff, that's that VNA X value that we've drawn in our diagram. OK. So that's what we're looking for. And we have four possible answer choices all in meters per second. Option A is 15. Option B is 25. Option C is 30 and option D is 20. Now we're told there's no air resistance. and what that means is that we're gonna have no horizontal acceleration. OK? So the initial horizontal velocity is gonna be equal to the final horizontal velocity. We're just gonna call both of them V X for simplicity. OK. So that's what we're looking for now is V X. But let's write out the variables we have in the horizontal direction because that's what we're interested in. And so in the X direction, we have that velocity of V X. What we're looking for, we have the horizontal displacement delta X which is 30 m to that horizontal distance traveled. We know that the time this takes is 1.5 seconds. And again, because we have no acceleration, we only have these three variables. And the equation that governs this motion recall is gonna be that V X is equal to delta X divided by T and the velocity is equal to the displacement divided by time. Now, we know delta X and T, we can substitute those values and solve for V X forget that this is equal to 30 m divided by 1. seconds. We can simplify this to 20 m per second. OK. So that's the answer. We were looking for the speed of the car as it takes off from the clip is gonna be m per second, which corresponds with answer choice D thanks everyone for watching. I hope this video helped see you in the next one.