{"id":2039,"date":"2021-06-22T03:46:15","date_gmt":"2021-06-21T22:16:15","guid":{"rendered":"https:\/\/acejee.com\/blog\/?p=2039"},"modified":"2024-09-30T08:11:10","modified_gmt":"2024-09-30T02:41:10","slug":"horizontal-range-of-projectile","status":"publish","type":"post","link":"https:\/\/acejee.com\/blog\/horizontal-range-of-projectile\/","title":{"rendered":"Horizontal Range of Projectile"},"content":{"rendered":"\t\t
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Horizontal range of a projectile is the horizontal distance travelled by the projectile between launch and the landing points.<\/p>

We will begin with an expression for the range for a projectile, projected at an angle $\\theta$ on a level ground meaning launch and landing points are at the same height.\u00a0<\/p>

Following this we will find out the maximum range of the projectile for a given initial velocity and as we will see it is maximum when angle of projection $\\theta = 45^\\circ$.\u00a0<\/p>

Afterwards we will see that range is same for $\\theta$ and $90^\\circ – \\theta$.\u00a0<\/p>

In the end we will find out range when ground is not level meaning landing point is at a different height than the launch point, say in case of horizontal projectile motion<\/p>

So, let’s start with the expression for the range<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Range of Projectile, $R$<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Horizontal\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Horizontal distance covered by the projectile by the time it returns to the ground is $=u_x \\times$ Time of Flight, $T$, where $u_x = u \\cos \\theta$ is the horizontal component of velocity which does NOT change during the flight as there is no acceleration along the horizontal direction<\/p>

Time of Flight, $T$, time over which $y$ displacement becomes zero is given by<\/p>

$0 = u_y T – \\cfrac{1}{2} g T^2$ or $T =$ $\\cfrac{2 u_y}{g}$\u00a0$=\\cfrac{2 u \\sin \\theta}{g}$<\/span><\/p>

which means, Range, $R$ $=\\cfrac{2 u^2 \\sin \\theta \\cos \\theta}{g}$\u00a0<\/span>$=\\cfrac{u^2 \\sin 2\\theta}{g}$<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Maximum Range of Projectile<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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