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

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Answer 1 · 2017-12-27T08:07:11

The peak height is $=3.27m$

Resolving in the vertical direction $uarr^+$

The initial velocity is $u=16sin(1/6pi)ms^-1$

Let the peak height be $=hm$

At the greatest heignt $v=0ms^-1$

The acceleration due to gravity is $g=9.8ms^-2$

Applying the equation of motion

$v^2=u^2+2as=u^2-2gh$

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

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