If a projectile is shot at an angle of $\frac{π}{6}$ $pi/6$ and at a velocity of $28 \frac{m}{s}$ $28 m/s$ , 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 $28 \frac{m}{s}$ $28 m/s$ , when will it reach its maximum height??

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Answer 1 · 2016-04-08T12:19:59

$\frac{10}{7} s$ $10/7s$

Explanation:

$t = \frac{u sin α}{g} = \frac{28 \times sin (\frac{π}{6})}{9.8} = \frac{28 \times \frac{1}{2}}{9.8} = \frac{140}{98} = \frac{10}{7} s$ $t =(usinalpha)/g=(28xxsin(pi/6))/9.8=(28xx1/2)/9.8=140/98=10/7s$

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If a projectile is shot at an angle of $\frac{π}{6}$ $pi/6$ and at a velocity of $28 \frac{m}{s}$ $28 m/s$ , 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 pi/6π6pi/6 and at a velocity of 28 m/s28ms28 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 $28 \frac{m}{s}$ $28 m/s$ , when will it reach its maximum height??