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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??

1 Answer

Jan 6, 2016

#h=6.747m#

Explanation:

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#

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