A projectile is shot at a velocity of $2 \frac{m}{s}$ $2 m/s$ and an angle of $\frac{π}{6}$ $pi/6$ . What is the projectile's peak height?

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A projectile is shot at a velocity of $2 \frac{m}{s}$ $2 m/s$ and an angle of $\frac{π}{6}$ $pi/6$ . What is the projectile's peak height?

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Answer 1 · 2016-03-01T10:53:09

$h_{p} ≅ 0, 051 m$ $h_p~=0,051 m$

Explanation:

$The projectile's peak height can be calculated using the formula:$ $"The projectile's peak height can be calculated using the formula:"$
$h_{p} = \frac{v_{i}^{2} \cdot {sin}^{2} α}{2 \cdot g}$ $h_p=(v_i^2*sin^2 alpha)/(2*g)$
$v_{i} :initial velocity of object$ $v_i" :initial velocity of object"$
$α :angle of projectile$ $alpha" :angle of projectile"$
$g : acceleration of gravity$ $g" : acceleration of gravity"$
$α = \frac{π}{6} \cdot \frac{180}{π} = 30^{o}$ $alpha=pi/6*180/pi=30^o$
$sin 30 = 0, 5$ $sin 30=0,5$
${sin}^{2} 30 = 0, 25$ $sin^2 30=0,25$
$h_{p} = \frac{2^{2} \cdot 0, 25}{2 \cdot 9, 81}$ $h_p=(2^2*0,25)/(2*9,81)$
$h_{p} = \frac{2 \cdot 0, 25}{9, 81}$ $h_p=(2*0,25)/(9,81)$
$h_{p} ≅ 0, 051 m$ $h_p~=0,051 m$

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A projectile is shot at a velocity of $2 \frac{m}{s}$ $2 m/s$ and an angle of $\frac{π}{6}$ $pi/6$ . What is the projectile's peak height?

1 Answer

Explanation:

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