A frog leaves the ground with a speed of 15 m/s and stays in the air for 2.0 seconds. At what angle did the frog jump?

Hey, guys. Let's get started with this problem. Hopefully, you got a chance to try it on your own. So we got a frog leaping with some speed and we're told it hits the ground 2 seconds later. We're going to calculate the angle that the frog jumps. So before we get started, I just wanna draw a quick sketch. It's always useful to do this. So the frog's gonna jump like this and it's gonna later return to the ground, which means we know we're dealing with a symmetrical launch problem, which is great because we can use all of our special properties and our equations to figure out the answer here in the target variable. So what are we actually looking for? We're looking for an angle, which is theta and we're looking for the angle that the frog jumped at. So before we get started, we're actually just gonna go ahead and label all the points of interest. So in the x axis, we'd go there and the y axis it would look like this and our points of interest are a, b, just the maximum height, and c when we hit the ground again. So if we draw our pass in the x and y, it would look like this. So what we're looking for here is we have the launch velocity. We have v naughts. That's 15 meters per second. What we don't know is we don't know the angle that this frog is going to jump at. So what else we told in the problem here, there's only one piece of information that we have, which is the time. So if we look through our special equations here, the reason that they're really useful is because they really just relate 3 variables or 4 variables. You have the time, which is related to the launch velocity and the angle, and you also have the range, which is also just equal to or also, proportional to the launch velocity and the angle. So if you know any 2 of those 3 variables or those 4 variables, you could always solve for the other one. So this is really useful because if we're trying to solve for time or for the angle here, the other piece of information that we have is we have the total time of flight. We know that this frog is gonna jump and it's gonna hit the ground 2 seconds later. So what that means is that t a c or delta t from a to c is equal to 2 seconds. So we have delta T from a to c, which is t symmetrical. So we have that. And we also have the launch velocity, which means we can figure out what the theta angle is just by using our total time of flight equation. So we're gonna be looking at the interval from a to c, and so our total time flight equation t a c is gonna be 2 times v naught sine theta over g. So if if we have everything, except for theta, we can just go ahead and plug everything in. Right? So TAC is gonna be 2 and we got 2 times 15, that's the launch velocity, times the sine of theta over 9.8. So, if we actually go ahead and rearrange, move the 9.8 over to the other side, we're gonna get 2 times 9.8 divided by and then 2 times 15. And that's gonna equal sine of theta a. So what happens is you're just gonna get a number out of this. And so if we wanted to get theta a, what happens is we've got this sign that's out in front. How do we get rid of it? Well, we just take the inverse sign. So we're gonna take the inverse sign to basically get rid of the sign that's wrapping up that theta a, and the number that you get when you do this is gonna be 0.65. So you've take the inverse sign and make sure that your calculator is in degrees mode. What you're gonna get is you're gonna get 40.8 degrees, and so that is the launch angle of the frog. So that's how these symmetrical launch problems and the equations are super useful because they really just make our symmetrical launches very, very simple and straight straightforward. Alright, guys. That's it for this one. Let me know if you have

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A frog leaves the ground with a speed of 15 m/s and stays in the air for 2.0 seconds. At what angle did the frog jump?

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