If a projectile is shot at an angle of $\frac{2 π}{3}$ $(2pi)/3$ and at a velocity of $64 \frac{m}{s}$ $64 m/s$ , 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 $64 \frac{m}{s}$ $64 m/s$ , when will it reach its maximum height?

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Answer 1 · 2016-03-11T03:48:27

$\approx 5.54 s$ $~~5.54s$

Explanation:

velocity of projection, $u = 64 m s^{- 1}$ $u=64ms^-1$
angle of projection, $α = 2 \frac{π}{3}$ $alpha=2pi/3$
if time of reaching maximum height be t
then it will have zero velocity at the peak.
So $0 = u \cdot sin α - g \cdot t$ $0=u*sinalpha- g*t$
$\Rightarrow t = u \cdot \frac{sin α}{g} = 64 \cdot \frac{sin (2 \frac{π}{3})}{10} = 6.4 \cdot \frac{\sqrt{3}}{2} = 3.2 \cdot \sqrt{3} m \approx 5.54 s$ $=>t=u*sinalpha/ g=64*sin(2pi/3)/10=6.4*sqrt3/2=3.2*sqrt3m~~5.54s$

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If a projectile is shot at an angle of $\frac{2 π}{3}$ $(2pi)/3$ and at a velocity of $64 \frac{m}{s}$ $64 m/s$ , 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 $64 \frac{m}{s}$ $64 m/s$ , when will it reach its maximum height?