 On a calm summer morning, you and your friends finally get a chance to go on a hot air balloon trip. You all get on board and rise up to an altitude of 300 meters. From there, at a constant height, a wind pushes you at a speed of 20 meters per second relative to the ground. You decide the moment needs a selfie, so you get your phone out, but in the excitement of the moment, you drop it over the edge. Where should you look for your phone on the ground? Secondly, once you know where to look for your phone, is it even worth bothering or will it be smashed to pieces? You quickly borrow your friend's phone to check the manufacturer's website. It turns out your phone is rated to survive impacts up to 50 meters per second. Is your phone likely to have survived the fall? Let's start with the first problem. We know that this is a projectile motion problem since we have an object moving under the influence of gravity only. We know that the phone will end up on the ground, so we're looking to find the distance that the phone travels from the hot air balloon. Now, we know the initial horizontal speed of the hot air balloon, vxi, and there are no horizontal forces, so we know that the horizontal velocity is constant, and the distance the phone travels horizontally is given by x is equal to vxi times t, where t is the time the phone takes to hit the ground. Now for t, we know the total displacement is 300 meters, the initial vertical velocity is 0 meters per second, and the acceleration is 9.8 meters per second squared. So we can use our projectile motion equation to find y equal to 300 meters. We get y is equal to half gt squared. We know that both y and g are acting downwards, and so they're acting in the same direction, and we can keep both as positive. Plugging in our numbers, we get a time to hit the ground of 7.8 seconds. We can now plug this into our equation for x. We find that you should look for your phone at a horizontal distance of 156 meters from where you were when you dropped the phone. Great, so we've worked out where you should look for your phone, but is it actually bothering to look, or will it be smashed? For this problem, we need to know the speed at which the phone is travelling. To do this, we're going to use vector addition, and sum both the horizontal and vertical velocities. We already know the horizontal velocity, because there are no horizontal forces. The final horizontal velocity is equal to the initial horizontal velocity of 20 meters per second. We can calculate the final vertical velocity from our projectile motion equations. The initial vertical velocity is zero, so the final vertical velocity, this gives us an answer of 77 meters per second. We can now sum our velocities using a vector diagram, and we find that the final velocity is equal to the square root of the vertical velocity squared, plus the horizontal velocity squared. Plugging in our numbers, we get a final velocity of 80 meters per second. Given that the phone is only raised at impacts up to 50 meters per second, it was probably smashed to bits when it hit the ground. Sorry to break the news.