A projectile is shot at a velocity of $8 \frac{m}{s}$ $8 m/s$ 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 $8 \frac{m}{s}$ $8 m/s$ and an angle of $\frac{π}{6}$ $pi/6$ . What is the projectile's peak height?

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Answer 1 · 2016-12-17T19:09:03

Equations of motion are used to find an answer of 0.816 m

Explanation:

Use the equation of motion that reads:

$v^{2}$ $v^2$ = $u^{2}$ $u^2$ - $2 g y$ $2gy$

Where $v$ $v$ is the final vertical velocity, $u$ $u$ is the initial vertical velocity and $y$ $y$ is the change in altitude (the vertical displacement) of the projectile.

Now, the launch angle is $π$ $pi$ /6, so $u$ $u$ = 8 sin $\frac{π}{6}$ $pi/6$ = 8 (0.5) = 4.0 $\frac{m}{s}$ $m/s$

"Reaches its greatest height" means $v$ $v$ = 0, so putting it together, we get

0 = ${4.0}^{2}$ $4.0^2$ - 2 (9.8) $y$ $y$

$y$ $y$ = $\frac{16}{19.6}$ $16/ 19.6$ = 0.816 m

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A projectile is shot at a velocity of $8 \frac{m}{s}$ $8 m/s$ and an angle of $\frac{π}{6}$ $pi/6$ . What is the projectile's peak height?

1 Answer

Explanation:

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A projectile is shot at a velocity of $8 \frac{m}{s}$ $8 m/s$ and an angle of $\frac{π}{6}$ $pi/6$ . What is the projectile's peak height?