[amp_mcq option1=”36\u00c2\u00b0” option2=”45\u00c2\u00b0” option3=”56\u00c2\u00b0” option4=”76\u00c2\u00b0” correct=”option2″]<\/p>\n
The correct answer is $\\boxed{\\text{B}}$.<\/p>\n
The horizontal range of a projectile is the distance it travels horizontally after being launched. The maximum height of a projectile is the highest point it reaches after being launched. The angle of projection is the angle between the initial velocity vector of the projectile and the horizontal.<\/p>\n
The horizontal range and maximum height of a projectile are equal when the angle of projection is $\\boxed{45^\\circ}$. This is because the horizontal range is given by the formula<\/p>\n
$$R = \\frac{v^2 \\sin(2 \\theta)}{g}$$<\/p>\n
where $v$ is the initial velocity of the projectile, $\\theta$ is the angle of projection, and $g$ is the acceleration due to gravity. The maximum height is given by the formula<\/p>\n
$$H = \\frac{v^2 \\sin^2(\\theta)}{2g}$$<\/p>\n
When $\\theta = 45^\\circ$, $\\sin(2 \\theta) = 1$ and $\\sin^2(\\theta) = \\frac{1}{2}$. Substituting these values into the formulas for $R$ and $H$, we get<\/p>\n
$$R = \\frac{v^2}{g}$$<\/p>\n
$$H = \\frac{v^2}{4g}$$<\/p>\n
Therefore, the horizontal range and maximum height of a projectile are equal when the angle of projection is $45^\\circ$.<\/p>\n
The other options are incorrect because they are not angles that make the horizontal range and maximum height of a projectile equal.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”36\u00c2\u00b0” option2=”45\u00c2\u00b0” option3=”56\u00c2\u00b0” option4=”76\u00c2\u00b0” correct=”option2″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[645],"tags":[],"class_list":["post-6883","post","type-post","status-publish","format-standard","hentry","category-applied-mechanics-and-graphic-statics","no-featured-image-padding"],"yoast_head":"\n