A projectile is shot at a velocity of $27 m/s$ and an angle of $pi/12$ . What is the projectile's maximum height?

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Answer 1 · 2017-04-28T06:32:12

The maximum height is $=2.49m$

Resolving in the vertical direction $uarr^+$

initial velocity is $u_y=vsintheta=27*sin(1/12pi)$

At the maximum height, $v=0$

We apply the equation of motion

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

to calculate the greatest height

$0=(27sin(1/12pi)^2-2g*h$

$h=1/(2g)*(27sin(1/12pi))^2$

$=2.49m$

A projectile is shot at a velocity of 27 m/s27 m/s and an angle of pi/12 pi/12 . What is the projectile's maximum height?