{"id":2068,"date":"2021-07-01T13:20:11","date_gmt":"2021-07-01T07:50:11","guid":{"rendered":"https:\/\/acejee.com\/blog\/?p=2068"},"modified":"2024-09-30T08:11:08","modified_gmt":"2024-09-30T02:41:08","slug":"velocity-of-projectile-at-any-instant","status":"publish","type":"post","link":"https:\/\/acejee.com\/blog\/velocity-of-projectile-at-any-instant\/","title":{"rendered":"Velocity of Projectile at Any Instant"},"content":{"rendered":"\t\t
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Velocity of projectile at any instant can be written, in the vector form, as $\\overrightarrow{v}(t) =$ $v_x \\hat{i} +$ $v_y \\hat{j}$, where $v_x$ is the $x$ component of velocity vector $\\overrightarrow{v}(t)$ at time $t$ and $v_y$ is its $y$ component at that time.<\/p>

So let’s see, what will this expression be, in terms of initial speed $u$ and projection angle $\\theta$.<\/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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Velocity of projectile at any instant<\/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\"Velocity\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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Now, $v_x$, $x$ component of velocity vector $\\overrightarrow{v}(t)$ is simply $=u_x=$ $u \\cos \\theta$, as there is no acceleration in the horizontal direction.<\/p>

As for $v_y$, $y$ component of velocity vector $\\overrightarrow{v}(t)$, $v_y =$ $u_y – gt$ $=u \\sin \\theta – gt$<\/p>

So, velocity vector $\\overrightarrow{v}(t)$ will simply be $=u \\cos \\theta \\hat{i} +$ $(u \\sin \\theta – gt)$ $\\hat{j}$<\/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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Velocity of projectile at 'half' of maximum height<\/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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Now, let’s determine the expression for velocity vector $\\overrightarrow{v} =$ $v_x \\hat{i} +$ $v_y \\hat{j}$ at height $y$.<\/p>

So, at height $h$, $v_x$, $x$ component of velocity vector $\\overrightarrow{v}$ is simply $=u_x=$ $u \\cos \\theta$, as there is no acceleration in the horizontal direction.<\/p>

As for $v_y$, recall that $v_y^2 = u_y^2 – 2gy$ $= u^2 \\sin^2 \\theta – 2gy$, so $v_y=$ $\\pm \\sqrt{u^2 \\sin^2 \\theta – 2gy}$. Note that projectile is at height $h$ twice during it’s slight (once on the way up and once on the way down and hence $\\pm$)<\/p>

$\\therefore$ velocity vector $\\overrightarrow{v}(y)$ will simply be $=u \\cos \\theta \\hat{i} $ $\\pm \\sqrt{u^2 \\sin^2 \\theta – 2gy} \\hat{j}$<\/p>

So, what will the velocity of projectile be at ‘half’ the maximum height, i.e. at $y = \\cfrac{H}{2}$, where maximum height $H=\\cfrac{u^2 \\sin^2 \\theta}{2g}$<\/p>

Well, that will simply be $=u \\cos \\theta \\hat{i} $ $\\pm \\cfrac{u \\sin \\theta}{\\sqrt{2}} \\hat{j}$ or speed at that point will be $\\sqrt{\\cfrac{1+u^2 \\cos^2 \\theta}{2}}$.<\/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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The speed of a projectile at its maximum height<\/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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Well at maximum height $H$, $y$ component of velocity vector becomes zero, so $\\overrightarrow{v}(H)=u \\cos \\theta \\hat{i}$.<\/p>

With that, now let’s determine the average velocity vector of a projectile for two scenarios: 1) over a symmetric part of the journey, meaning, between two points at same height 2) over non-symmetric part of the journey, meaning average velocity between two points that are at different heights.<\/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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Average velocity of a particle in projectile motion (symmetric)<\/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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So, starting with average velocity over a symmetric part of the journey, meaning, average velocity between two points located at the same height, say for example, landing and launch points, at ground level.<\/p>

Recall that average velocity is defined as $=\\cfrac{Net \\ Displacement}{Time \\ Duration}$ $=\\cfrac{\\overrightarrow{r}_f – \\overrightarrow{r}_i}{\\Delta t}$<\/p>

For points located at same height,<\/p>

$\\overrightarrow{r}_f – \\overrightarrow{r}_i=$ $(x_f – x_i) \\hat{i}$ and $\\Delta t$ between $x_f$ and $x_i$ is simply $=\\cfrac{x_f – x_i}{u \\cos \\theta}$<\/p>

so, $\\overrightarrow{v}_{avg} = u \\cos \\theta \\hat{i}$<\/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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Average velocity over a non-symmetric part of the journey<\/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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As for average velocity over a non symmetric part of the flight, meaning, average velocity between two points at different heights,\u00a0 it can be calculated using the definition\u00a0$\\overrightarrow{v}_{avg}\u00a0=$ $\\cfrac{\\overrightarrow{r}_f – \\overrightarrow{r}_i}{\\Delta t}$.<\/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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When is the velocity at the maximum height of a projectile, half its initial velocity?<\/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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Speed at the maximum height of a projectile, $u \\cos ]theta$, will be half its initial speed, $u$, when $\\cos \\theta = \\cfrac{1}{2}$ or $\\theta = 60^\\circ$.<\/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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Projectile Motion | Important Questions | JEE PYQs<\/p>