In a projectile motion, the horizontal range and the maximum height attained by

Question

In a projectile motion, the horizontal range and the maximum height attained by the projectile are equal. a) What is the launch angle? b) If everything else stays the same, how should the launch angle, ?0, of a projectile be changed for the range of the projectile to be halved?

Explanation / Answer

Let u = initial velocity theta = angle with respect to the horizontal => Range = u^2 sin(2*theta)/g Max Height = u^2 sin^2(theta)/2g also for Range = Maximum height => u^2 sin(2*theta)/g = u^2 sin^2(theta)/2g => 2sin(theta) cos(theta) = sin^2(theta)/2 => tan(theta) = 4 or theta = tan^-1 (4) = 75.96 degrees -------(a) for range to be halved, we have new range = u^2 sin(2*alpha)/g --------(alpha being the new angle) => u^2 sin(2*alpha)/g = u^2 sin(2*theta)/2g => sin(2*alpha) = 1/2 * sin(2*theta) => alpha = 1/2 * sin^-1(1/2 * sin(2*theta)) => alpha = 6.804 degrees

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