If a projectile is shot at an angle of $\frac{7 π}{12}$ $(7pi)/12$ 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 $\frac{7 π}{12}$ $(7pi)/12$ 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 · 2016-03-14T10:06:40

$t_{e} = 0, 098 s$ $t_e=0,098 s$

Explanation:

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$α = 7 \frac{π}{12}$ $alpha=7pi/12$
$v_{i} = 1 \frac{m}{s}$ $v_i=1 m/s$
$t_{e} = v_{i} \cdot \frac{sin α}{g} elapsed time to the maximum height$ $t_e=v_i*sin alpha/g" elapsed time to the maximum height"$
$t_{e} = \frac{1 \cdot sin (7 \frac{π}{12})}{9, 81}$ $t_e=(1*sin(7pi/12))/(9,81)$
$t_{e} = 0, 098 s$ $t_e=0,098 s$

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If a projectile is shot at an angle of $\frac{7 π}{12}$ $(7pi)/12$ and at a velocity of $1 \frac{m}{s}$ $1 m/s$ , 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 $\frac{7 π}{12}$ $(7pi)/12$ and at a velocity of $1 \frac{m}{s}$ $1 m/s$ , when will it reach its maximum height?