If a projectile is shot at an angle of $\frac{7 π}{12}$ $(7pi)/12$ and at a velocity of $2 \frac{m}{s}$ $2 m/s$ , 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 $2 \frac{m}{s}$ $2 m/s$ , when will it reach its maximum height?

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Answer 1 · 2016-04-23T11:51:55

Time $t = \frac{5 \sqrt{6} + 5 \sqrt{2}}{98} = 0.1971277197$ $t=(5sqrt6+5sqrt2)/98=0.1971277197" "$ second

Explanation:

For the vertical displacement $y$ $y$

$y = v_{0} sin θ \cdot t + \frac{1}{2} \cdot g \cdot t^{2}$ $y=v_0 sin theta * t+1/2*g*t^2$

We maximize displacement $y$ $y$ with respect to $t$ $t$

$\frac{d y}{d t} = v_{0} sin θ \cdot \frac{d t}{d t} + \frac{1}{2} \cdot g \cdot 2 \cdot t^{2 - 1} \cdot \frac{d t}{d t}$ $dy/dt=v_0 sin theta* dt/dt+1/2*g*2*t^(2-1)*dt/dt$

$\frac{d y}{d t} = v_{0} sin θ + g \cdot t$ $dy/dt=v_0 sin theta+g*t$

set $\frac{d y}{d t} = 0$ $dy/dt=0$ then solve for $t$ $t$

$v_{0} sin θ + g \cdot t = 0$ $v_0 sin theta+g*t=0$

$t = \frac{- v_{0} sin θ}{g}$ $t=(-v_0 sin theta)/g$

$t = \frac{- 2 \cdot sin (\frac{7 π}{12})}{- 9.8}$ $t=(-2*sin ((7pi)/12))/(-9.8)$

Note: $sin (\frac{7 π}{12}) = sin (\frac{5 π}{12}) = \frac{\sqrt{6} + \sqrt{2}}{4}$ $sin ((7pi)/12)=sin ((5pi)/12)=(sqrt(6)+sqrt(2))/4$

$t = \frac{- 2 \cdot \frac{(\sqrt{6} + \sqrt{2})}{4}}{- 9.8}$ $t=(-2*((sqrt(6)+sqrt(2)))/4)/(-9.8)$

$t = \frac{5 \sqrt{6} + 5 \sqrt{2}}{98} = 0.1971277197$ $t=(5sqrt6+5sqrt2)/98=0.1971277197" "$ second

God bless....I hope the explanation is useful.

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If a projectile is shot at an angle of $\frac{7 π}{12}$ $(7pi)/12$ and at a velocity of $2 \frac{m}{s}$ $2 m/s$ , 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 $2 \frac{m}{s}$ $2 m/s$ , when will it reach its maximum height?