If a projectile is shot at an angle of $\frac{2 π}{3}$ $(2pi)/3$ and at a velocity of $1 \frac{m}{s}$ $1 m/s$ , when will it reach its maximum height?

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If a projectile is shot at an angle of $\frac{2 π}{3}$ $(2pi)/3$ and at a velocity of $1 \frac{m}{s}$ $1 m/s$ , when will it reach its maximum height?

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Answer 1 · 2017-07-21T17:24:07

$t = 0.0883$ $t = 0.0883$ $s$ $"s"$

Explanation:

We're asked to find the time $t$ $t$ when a particle reaches its maximum height, given its initial velocity.

To do this, we can use the equation

$v_{y} = v_{0} {sin α}_{0} - g t$ $v_y = v_0sinalpha_0 - g t$

where

$v_{y}$ $v_y$ is the velocity at time $t$ $t$ (this will be $0$ $0$ , since at a particle's maximum height its instantaneous $y$ $y$ -velocity is $0$ $0$ )
$v_{0}$ $v_0$ is the initial speed (give as $1$ $1$ $m/s$ $"m/s"$ )
$α_{0}$ $alpha_0$ is the launch angle (given as $\frac{2 π}{3}$ $(2pi)/3$ )
$g$ $g$ is the acceleration due to gravity near earth's surface ( $9.81$ $9.81$ ${m/s}^{2}$ $"m/s"^2$ )
$t$ $t$ is the time (what we're trying to find)

Plugging in known values, we have

$0 = (1 l m/s) sin (\frac{2 π}{3}) - (9.81 l {m/s}^{2}) t$ $0 = (1color(white)(l)"m/s")sin((2pi)/3) - (9.81color(white)(l)"m/s"^2)t$

$t = \frac{0.866 l m/s}{9.81 l {m/s}^{2}} = 0.0883$ $t = (0.866color(white)(l)"m/s")/(9.81color(white)(l)"m/s"^2) = color(red)(0.0883$ $s$ $color(red)("s"$

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If a projectile is shot at an angle of $\frac{2 π}{3}$ $(2pi)/3$ and at a velocity of $1 \frac{m}{s}$ $1 m/s$ , when will it reach its maximum height?

1 Answer

Explanation:

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If a projectile is shot at an angle of $\frac{2 π}{3}$ $(2pi)/3$ and at a velocity of $1 \frac{m}{s}$ $1 m/s$ , when will it reach its maximum height?