If a projectile is shot at an angle of $\frac{2 π}{3}$ $(2pi)/3$ and at a velocity of $19 \frac{m}{s}$ $19 m/s$ , 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 $19 \frac{m}{s}$ $19 m/s$ , when will it reach its maximum height?

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Answer 1 · 2017-07-01T17:09:43

$t = 1.68$ $t = 1.68$ $s$ $"s"$

Explanation:

We're asked to find the time when a projectile reaches its maximum height, given its initial velocity.

When the projectile is at its maximum height, its instantaneous $y$ $y$ -velocity is zero.

We can use the equation

$v_{y} = v_{0 y} - g t$ $v_y = v_(0y) - g t$

to find the time.

The initial $y$ $y$ -velocity $v_{0 y}$ $v_(0y)$ is given by

$v_{0 y} = v_{0} {sin α}_{0}$ $v_(0y) = v_0sinalpha_0$

$v_{0 y} = (19 l m/s) sin (\frac{2 π}{3}) = 16.5$ $v_(0y) = (19color(white)(l)"m/s")sin((2pi)/3) = color(red)(16.5$ $m/s$ $color(red)("m/s"$

And $g = 9.81$ $g = 9.81$ ${m/s}^{2}$ $"m/s"^2$

Plugging in known values, we have

$0 = 16.5 l m/s - (9.81 l {m/s}^{2}) t$ $0 = color(red)(16.5)color(white)(l)color(red)("m/s") - (9.81color(white)(l)"m/s"^2)t$

$t = \frac{16.5 l m/s}{9.81 l {m/s}^{2}} = 1.68$ $t = (color(red)(16.5)color(white)(l)color(red)("m/s"))/(9.81color(white)(l)"m/s"^2) = color(blue)(1.68$ $s$ $color(blue)("s"$

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If a projectile is shot at an angle of $\frac{2 π}{3}$ $(2pi)/3$ and at a velocity of $19 \frac{m}{s}$ $19 m/s$ , 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 $19 \frac{m}{s}$ $19 m/s$ , when will it reach its maximum height?