A projectile is shot at a velocity of $3 \frac{m}{s}$ $3 m/s$ and an angle of $\frac{π}{8}$ $pi/8$ . What is the projectile's peak height?

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A projectile is shot at a velocity of $3 \frac{m}{s}$ $3 m/s$ and an angle of $\frac{π}{8}$ $pi/8$ . What is the projectile's peak height?

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Answer 1 · 2016-02-03T18:49:12

$h_{p e a k} = 0, 00888$ $h_(peak)=0,00888$ $meters$ $"meters"$

Explanation:

$the formula needed to solve this problem is:$ $"the formula needed to solve this problem is:"$
$h_{p e a k} = (v_{i}^{2} \cdot \frac{{sin}^{2} θ}{2 \cdot g})$ $h_(peak)=(v_i^2*sin^2 theta/(2*g))$
$v_{i} = 3 \frac{m}{s}$ $v_i=3 m/s$
$theta =180/cancel(pi)*cancel(pi)/8$
$theta =180/8$
$sin theta=0,13917310096$
$sin^2 theta=0,0193691520308$
$h_(peak)=3^2*(0,0193691520308)/(2*9,81)$
$h_(peak)=9*(0,0193691520308)/(19,62)$
$h_(peak)=0,00888$ $"meters"$

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A projectile is shot at a velocity of $3 \frac{m}{s}$ $3 m/s$ and an angle of $\frac{π}{8}$ $pi/8$ . What is the projectile's peak height?

1 Answer

Explanation:

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