If a projectile is shot at an angle of $\frac{2 π}{3}$ $(2pi)/3$ and at a velocity of $26 \frac{m}{s}$ $26 m/s$ , 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 $26 \frac{m}{s}$ $26 m/s$ , when will it reach its maximum height?

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Answer 1 · 2016-04-04T11:09:51

$t_{m} = 2, 29 s$ $t_m=2,29 " s"$

Explanation:

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$given data:$ $"given data:"$
$initial velocity : v_{i} = 26 \frac{m}{s}$ $"initial velocity :"v_i=26 m/s$
$angle : α = \frac{2 π}{3}$ $"angle :"alpha=(2pi)/3$
$\frac{sin (2 π)}{3} = 0, 866$ $sin(2pi)/3=0,866$
$elapsed time to reach its maximum height : t_{m} = \frac{v_{i} \cdot sin α}{g}$ $"elapsed time to reach its maximum height :"t_m=(v_i*sin alpha)/g$
$g = 9, 81 \frac{m}{s^{2}}$ $g=9,81 m/s^2$

$t_{m} = \frac{26 \cdot 0, 866}{9, 81}$ $t_m=(26*0,866)/(9,81)$

$t_{m} = 2, 29 s$ $t_m=2,29 " s"$

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If a projectile is shot at an angle of $\frac{2 π}{3}$ $(2pi)/3$ and at a velocity of $26 \frac{m}{s}$ $26 m/s$ , when will it reach its maximum height?

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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 $26 \frac{m}{s}$ $26 m/s$ , when will it reach its maximum height?