A projectile is shot at a velocity of $2 \frac{m}{s}$ $2 m/s$ 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 $2 \frac{m}{s}$ $2 m/s$ and an angle of $\frac{π}{8}$ $pi/8$ . What is the projectile's peak height?

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Answer 1 · 2015-12-29T20:04:16

The projectile's peak height is $23.69 m$ $23.69m$

Explanation:

You have to consider the two component of motion
$s (x) = v t$ $s(x)=vt$ with $v=vcos22.5°$
$s(y)=vt-1/2at^2$ with $v=vsin22.5°$
You have to find the moment when the vertical speed is =0, that is the moment of the peak.
So $v=v-at=0$ , $t=g/v=12.82s$
Use this value in the equation of horizontal component
$s(x)=(vcos22.5°)*12.82=23.69m$

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A projectile is shot at a velocity of $2 \frac{m}{s}$ $2 m/s$ and an angle of $\frac{π}{8}$ $pi/8$ . What is the projectile's peak height?

1 Answer

Explanation:

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