The range R of a projectile fired from the origin over horizon...

Question

The range R of a projectile fired from the origin over horizontal ground is the distance from the origin to the point of impact. If the projectile is fired with an initial velocity at an angle with the horizontal, then in Chapter 13 we find that R-v_0^2/g(sin 2α) where g is the downward acceleration due to gravity. Find the angle α for which the range R is the largest possible.

Accepted Answer

Welcome back, everyone. In this problem, a golfer aims to hit a golf ball to achieve the greatest horizontal range. The ranger of a golf ball can be calculated using the formula R equals V0 square divided by Gs signed to theta, where V0 is the initial speed of the golf ball, G is the acceleration due to gravity, and the theta is the launch angle above the horizontal. What is the optimal angle theta for achieving the maximum range? A says it's 15 degrees, B 30 degrees, C 45 degrees, and the D 60 degrees. Now, to determine the optimal angle theta for the maximum range, basically we need to find the value of theta that maximizes the range formula. No. To maximize, OK, maximizing R equals V 2 divided by G sin to theta, OK. By looking at it, first we can tell that V0 square divided by G is a constant, assuming the initial speed and gravity don't change. Thus, the maximum range will occur when si in 2 theta is at its maximum value. Now we know that theater. is going to be between 0 and 90 degrees, OK? Or we could even say that it's between 0 and half of pi radiance, OK? I know in that case, what do we know about the sign function for this range? Well, we know the sine function reaches its maximum value of 1 when its argument equals 90 degrees or half of high radiance based on our range here, OK? So That means since this is true about theater, then signed to a theater. Is maximum, OK. When the argument of our assign function to theta is equal to 90 degrees, and if 2 theta equals 90 degrees, then theta must be equal to 45 degrees. Thus, this must be the angle for which it it gives us the maximum range. But let's not assume it. We need to verify if this would indeed result to a maximum range. To verify it, we can take the derivative of the range function with respect to theta and set it equal to 0. Then we can confirm if it's a maximum using the second derivative test. Now, if we do that, OK. Then we also know that at the mark at the critical point of the function rather. At the critical point of the function, the first derivative of R with respect to theta is going to be equal to 0. Now if we differentiate R with respect to theta, that means we are finding V not square divided by G multiplied by the derivative with respect to theta of assigned to the theta, OK. And that is gonna be equal to V0 square divided by G multiplied by 2 cost 2 theta. Now, if we set this equal to 0, OK, then this would imply that cost 2 theater. Would be equal to 0, and this occurs when 2 theta equals 90, so 2 theta would be equal to 90 degrees, which means that theta would be equal to 45 degrees. And know that we have our critical value which we we know that we would have found by reasoning it through. By the second derivative test, OK. If the second derivative is. Positive for that value, then we know that it's a minimum, but if it's negative for the value, then we know that it will be a maximum. OK. So by the second derivative test, OK, since the second derivative of R with respect to theta, OK, is going to be equal to, we, to find that we can differentiate our first derivative again. So it's gonna be equal to V 2 divided by G multiplied by the derivative with respect to theta of 2 + 2 theta, and that is gonna be equal to V 2 divided by G multiplied by -4 sin to theta, OK? And no, at theta equals to 45 degrees. Then we're going to have uh the sign of two theater. To be equal to the sine of 90, which equals 1 and thus if this expression sin to theta equals 1, then that means our second derivative. Is gonna be equal to square divided by G multiplied by -4 multiplied by 1, which is going to be a negative value, OK, regardless of what our initial speed is because remember our initial speed is squared, so it's gonna end up giving us a positive value anyways. So the second derivative is going to be negative and thus that tells us that theta equals 45 degrees is a maximum. So now we've also been able to test and prove that the optimal angle theater for achieving the maximum range is 45 degrees. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Projectile Motion

Trigonometric Functions

Optimization in Calculus

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