If a projectile is shot at an angle of $pi/6$ and at a velocity of $23 m/s$ , when will it reach its maximum height??

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Answer 1 · 2016-01-06T05:55:46

$h=6.747m$

Data:-
Angle $=theta=pi/6$
Initial Velocity $=$ Muzzle Velocity $=v_0=23m/s$
Acceleration due to gravity $=g=9.8m/s^2$
Maximum Height $=h=??$

Sol:-
We know that:
$h=(v_0^2sin^2theta)/(2g)$
$implies h=(23^2*sin^2(pi/6))/(2*9.8)=(529*0.25)/19.6=132.25/19.6=6.747m$
$implies h=6.747m$

If a projectile is shot at an angle of pi/6pi/6 and at a velocity of 23 m/s23 m/s, when will it reach its maximum height??