 In this video, we're going to work out lots of the information you might want to know about an object undergoing projectile motion. These example videos will be much more useful to you if you pause and have a go yourself before watching how it's done. A dolphin jumps out of the water. If the dolphin jumps at a speed of 5 meters per second and at an angle of 60 degrees to the horizontal, A, what is the initial vertical velocity of the dolphin? B, what is the initial horizontal velocity of the dolphin? C, at what distance does the dolphin re-enter the water? D, at what time does the dolphin re-enter the water? And E, what is the maximum height reached by the dolphin? We'll also look at a couple of questions about finding an unknown angle using projectile motion as follows. What angle to the horizontal should the dolphin jump? A, to reach a height of 1.1 meters, and B, to re-enter the water at the maximum distance. Firstly, we're going to calculate the initial vertical and horizontal velocities. We can draw a component diagram of the velocity, and we can see that we can calculate Vi x and Vi y from trigonometry. Now we know that initial velocity Vi is 5 meters per second, and we know that theta, the angle to the horizontal, is 60 degrees. So Vi x must be 5 on 2 meters per second, and Vi y must be 5 times root 3 on 2 meters per second. I've left it as an improper fraction, but you could also put sign of 60 degrees or cos of 60 degrees into your calculator and get a decimal number. Next up, we'll calculate the time of flight, or the time the dolphin re-enters the water. To solve this, we only need the equation for the vertical position. If the dolphin starts at time t equals 0, and height y equals 0, its height is just given by y equals g times t squared on 2, plus the initial y velocity times t. When y drops back down to 0, the dolphin will be back at the water surface, so all we need to do is plug in the height y equals 0 and solve for the time t. Don't forget that gravity is acting downwards, so g is negative 9.8 meters per second squared. So we end up with the total time of flight, tf, is equal to 0.884 seconds to 3 significant figures. Up next, we're going to find how far the dolphin jumped, or xf. This only depends on the horizontal velocity and the time the dolphin spent in the air. Since the horizontal velocity is constant, we just need to multiply the horizontal velocity, bix, by the time spent in the air, tf. If we plug in our previous values, this comes out 2.21 meters. Now we're going to look at the maximum height. The dolphin will reach its maximum height at tf divided by 2, halfway through the total time it spends in the air. This is because it will take the dolphin just as long to go up as it will to come back down. So now we can use the equation for the vertical position to find the position at the point in time halfway through the jump. And this vertical position will be the maximum height of the jump. Substituting in our values and cancelling our units, we find that the maximum height is equal to 0.957 meters to 3 significant figures. Finally, we're going to look at a couple of slightly trickier problems. This time, we're going to be solving for theta, given some other constraint. Firstly, what angle should the dolphin jump to jump to a height of 1.1 meters? Well, assuming that the dolphin's initial velocity is still 5 meters per second, we can use the same equation for the max height that we worked up before. So ymax is equal to viy squared on 2g. And we can relate the initial vertical velocity, viy, to the angle of launch from the velocity component triangle. So we know that viy is equal to vi times sine of theta. Rearranging for viy and then swapping viy for vi times sine theta, we get, now we want to get theta on its own. We divide both sides by vi and take the inverse sine of both sides. Plugging in all values and calculating, we get 68.2 degrees to 3 significant figures. Now we're going to figure out the angle to get the maximum range. This derivation is the most complex, so hang in there. Our general strategy to find the maximum range is to get an expression for the range, xmax, in terms of theta, and then figure out what value of theta will make xmax the greatest. xmax, the range, is just vix times t, the time of flight. Vix can be replaced with vi times cos theta, but what about t? Well, the time of flight can be found by setting height y equals 0 and then solving for t. The right-hand side can then be rewritten as... Then, since y equals 0, we can divide both sides by t and then rearrange the final equation to find that t equals 2 times vi times sine theta on g. Now we can put the time of flight t into our equation for x. We end up with x is equal to 2 times vi squared times cos theta times sine theta over g. So we want the value of the theta that makes x the biggest. It's not obvious what theta should be to make x as big as possible, but if we remember the trigonometry rules, we can rewrite 2 times sine theta times cos theta as sine of 2 theta. Making this replacement, the range simplifies down to x equals vi squared times sine of 2 theta over g. Since vi and g are fixed, the maximum x occurs when sine of 2 theta is maximized. The greatest value that sine can ever have is 1, which occurs when the argument is 90 degrees. This means that 2 theta is 90 degrees when sine is 1 and the range is maximized, which means that theta is equal to 45 degrees. So at 45 degrees, the range of the dolphins jump is the greatest it can be. We can take this a step further and work out a general formula for the maximum range of projectile motion, which is launched and lands at the same height. We already have an equation for our range x, which depends on vi theta and g. If we substitute theta equals 45 degrees into the equation, then sine of 2 theta is 1 and the maximum range is vi squared over g. And this is the maximum range for any projectile motion, which is launched from and lands at the same height. For this particular situation, we can work out dolphins' maximum range by substituting in vi, which is 5 meters per second, and g, which is 9.8 meters per second squared. This yields a maximum range of 2.55 meters. I covered a lot in this video, but by now you've seen most of the questions you might be asked about projectile motion calculations.