If a projectile is shot at an angle of $\frac{π}{6}$ $pi/6$ and at a velocity of $18 \frac{m}{s}$ $18 m/s$ , when will it reach its maximum height??

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If a projectile is shot at an angle of $\frac{π}{6}$ $pi/6$ and at a velocity of $18 \frac{m}{s}$ $18 m/s$ , when will it reach its maximum height??

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Answer 1 · 2016-03-30T13:04:13

Time of reaching at maximum height
$t = \frac{u sin α}{g} = \frac{18 \cdot sin (\frac{π}{6})}{9.8} = 0.91 s$ $t =(usinalpha)/g=(18*sin(pi/6))/9.8=0.91s$

Answer 2 · 2016-03-30T13:04:36

$t ≅ 0, 92 s$ $t~=0,92 " s"$

Explanation:

$g = 9, 82 \frac{m}{s^{2}}$ $g=9,82 " m/s^2$

$v_{i} = 18 \frac{m}{s}$ $v_i=18" "m/s$

$α = \frac{π}{6}$ $alpha=pi/6$

$sin α = 0, 5$ $sin alpha=0,5$

$t = \frac{v_{i} \cdot sin α}{g}$ $t=(v_i*sin alpha)/g$

$t = \frac{18 \cdot 0, 5}{9, 81}$ $t=(18*0,5)/(9,81)$

$t = \frac{9}{9, 81}$ $t=9/(9,81)$

$t ≅ 0, 92 s$ $t~=0,92 " s"$

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If a projectile is shot at an angle of $\frac{π}{6}$ $pi/6$ and at a velocity of $18 \frac{m}{s}$ $18 m/s$ , when will it reach its maximum height??

2 Answers

Explanation:

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If a projectile is shot at an angle of $\frac{π}{6}$ $pi/6$ and at a velocity of $18 \frac{m}{s}$ $18 m/s$ , when will it reach its maximum height??