 So let's do another projectile motion problem. This time we are given the initial velocity, we are given the initial angle, we are given the initial range, but we are wondering how long does it take until we reach the highest point and how high will the highest point actually be. Now what does it mean if we reach the highest point? Reaching the highest point means that the velocity in y direction must be zero. We are changing from a positive y into a negative y. So I can use this to write my velocity in y as a function of time is my wall velocity in y initial plus acceleration times time. Now we know that at the highest point the velocity will be zero. The initial we got is 40 meters per second times sine of 30. For the acceleration, assuming that we are on Earth, it's minus 9.8 meters per second square. Let's be a bit lazy and go with minus 10 meters per second square. And then times the time, that is the unknown variable. So if I rearrange this, I get 10 meters per second square times time gives me 40 times sine of 30, that's 20 meters per second. So time is 20 meters per second over 10 meters per second square, which gives me 2 meters cancel meters, 1 second over 2 seconds. Which of course in this case I could have gotten much easier because I know that the total flight time was 4 seconds. If we start from the same height as we are landing, the whole thing must be symmetrical and the time must be 2 seconds to get to the highest point. Now what is the highest point? So we figured out t to h max is 2 seconds. Now how high is the highest point? For that we're going to use this equation where we're simply going to plug in the time that we just figured out. So we're going to be using this equation with information that we have. So as in y direction, as a function of time, so in this case at 2 seconds must be v y initial times t plus 1 half at squared. So in our case at t equals 2 seconds, which is our highest point, I get 40 meters per second times sine of 30 times 2 seconds plus 1 half times 9.8. We're going to be lazy again. We're going to use 10, so minus 5 meters per second square times the time 2 square seconds square. And all we have to do is plug in the numbers. So 40 meters per second times sine of 30 times 2 seconds that gives me 40 meters minus 5 meters per second square, the second square cancel second square, 2 squared is 4 minus 5 times 2 is minus 20 meters and 5 times 4 centimeters gives you an answer of height of 20 meters. As you've seen, often it is not necessary to use the range equation or the trajectory equation. The range equation as a function of time for many problems is just enough. You plug in what you know in one direction, figure out the time used to time in the other direction to calculate whatever is needed.