(II) At what projection angle will the range of a projectile e...

Question

(II) At what projection angle will the range of a projectile equal its maximum height?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, find the angle at which a ball should be thrown so that its horizontal range matches half of its maximum height. So that's our end goals. We're trying to figure out what the angle has to be in order for a ball to be thrown, such that its horizontal range will match half of its maximum height. So that's what we're trying to solve for at the end, our end goal. Ultimately, our final answer is we're trying to figure out what this angle is. Awesome. So with that in mind, let's look at our multiple choice answers and read them off to see what our final answer might be. And let's also note that they're all in units of degrees. So A is 71 B is 76 C is 83 and D is 90. OK. So first off, let us choose the origin point to be where the ball is thrown upwards. And let us choose the upwards direction to be the positive Y direction. Let us also name the initial velocity of the projectile to be V zero. And let us name the launching angle to be theta zero. So now we need to recall and use the uniform accelerated motion equations to help us sol for the maximum height value. So let's recall that VY squared is equal to V zero, Y squared plus two multiplied by A Y multiplied by Y minus Y zero. And let's call this equation one. So let us note that Vy is equal to the Y component of velocity and that V zero Y is the Y component of the initial launch velocity which is equal to. So let us note here that V zero Y is equal to V zero multiplied by sine of theta zero. And let us also note that Vy is equal to zero meters per second when it's at its maximum height mean when you think about it. So this ball is thrown up into the air. So at one point at, for a split second, it's gonna be at 0 m per second. So it's not gonna be moving at its maximum height because it'll throw it up in the air. And for a split second, it'll be not moving before it starts falling back to earth. OK. And let's also note that A Y is the Y component of acceleration and this is equal to negative G where G denotes gravity. So Therefore, this value is equal to negative 9.8 m per second squared and Y equals Y max which is the maximum height reached. And Y zero is the initial position of the ball. So Y zero will be equal to 0 m because initially the ball isn't moving, it's at rest. So now we can substitute in all of our known variables into equation one and we will get that 0 m per second squared is equal to V zero multiplied by sign up beta zero. Which here, I'll do a little note here because we forgot to say that Y equals Y max. So we don't forget. Awesome. So with that in mind, so now going back to our equation plugging in all of our known variables. So Vy squared is 0 m per second squared and it's equal to V zero multiplied by sine of theta zero minus two, multiplied by G multiplied by Y max. OK. So Y max, so now we need to rearrange this equation to isolate and sulfur Y max. And let's call this equation two. So when we rearrange it to isolate and sulfur Y max by itself, we will get that Y max is equal to V zero multiplied by sine of theta zero squared, all divided by two multiplied by G. OK. So now at this point, we need to recall and use the equation for the horizontal range of the ball which states that R and let's call this equation three. So the range R is equal to V zero squared multiplied by sine of two, multiplied by theta zero all divided by G where just to reiterate G is the, is the value for gravity. So now at this point, we need to find the angle at which. So let's make a little note here. So we need to find the angle where R equals Y max divided by two. So this is the angle. So let's let's put a little note here. So angle at and then arrow, so angle at R equals Y max divided by two. So therefore, we can write that drum roll that V zero squared multiplied by sine of two multiplied by theta zero, all divided by G is equal to V zero multiplied by sine of theta zero squared divided by four multiplied by G. We could take it even further and write that sine of two multiplied by theta zero is equal to sign of theta zero squared divided by four. So now we need to recall and apply the trigonometric identity. And let's write this in blue. So we need to recall this Trigon Mery or this trig principle I should say which states that sine of two multiplied by theta is equal to two multiplied by sine of theta multiplied by cosine of theta. OK. Using this trig identity, we could write that two multiplied by sine theta zero multiplied by cosine of theta zero is equal to sign of theta zero to the power of two all divided by four, which we can simplify to eight multiplied by cosine of theta zero, which is equal to sine of theta zero. Therefore, we can simplify once again and write that tangent of theta zero is equal to eight. And finally, we can solve for theta zero, which is our final answer. So to get theta zero all by itself, we need to say that theta zero is equal to inverse tangent of eight. So when we plug that into our calculator, we will find that it's equal to 82.87 degrees, which is approximately equal to 83 degrees when we round to the nearest whole number. And that's our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter C 83 degrees. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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(II) At what projection angle will the range of a projectile equal its maximum height?

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