A projectile is shot at a velocity of $16 \frac{m}{s}$ $16 m/s$ and an angle of $\frac{π}{6}$ $pi/6$ . What is the projectile's peak height?

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Answer 1 · 2017-12-27T08:07:11

The peak height is $= 3.27 m$ $=3.27m$

Resolving in the vertical direction ${↑ ⏐ ⏐}^{+}$ $uarr^+$

The initial velocity is $u = 16 sin (\frac{1}{6} π) m s^{- 1}$ $u=16sin(1/6pi)ms^-1$

Let the peak height be $= h m$ $=hm$

At the greatest heignt $v = 0 m s^{- 1}$ $v=0ms^-1$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

Applying the equation of motion

$v^{2} = u^{2} + 2 a s = u^{2} - 2 g h$ $v^2=u^2+2as=u^2-2gh$

$h = \frac{u^{2} - v^{2}}{2 g} = \frac{{(16 sin (\frac{1}{6} π))}^{2} - 0}{2 g} = 3.27 m$ $h=(u^2-v^2)/(2g)=((16sin(1/6pi))^2-0)/(2g)=3.27m$

A projectile is shot at a velocity of 16 m/s16ms 16 m/s and an angle of pi/6 π6pi/6 . What is the projectile's peak height?