If a projectile is shot at an angle of #pi/6# and at a velocity of #32 m/s#, when will it reach its maximum height??

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If a projectile is shot at an angle of #pi/6# and at a velocity of #32 m/s#, when will it reach its maximum height??

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Answer 1 · 2016-02-08T11:04:00

#h_m=13,048 m " maximum height"#

Explanation:

#theta =cancel(pi)/6* 180/cancel(pi)=30^o#
#t=w_i*sin theta/g#
#t=32*sin 30#
#t=32*0,5=16 sec" time to maximum height"#
#h_m=(v_i^2*sin^2 30)/(2*g)#
#h_m=(32^2* 0,5^2)/(2*9,81)=(1024*0,25)/(19,62)#
#h_m=256/(19,62)#
#h_m=13,048 m " maximum height"#

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If a projectile is shot at an angle of #pi/6# and at a velocity of #32 m/s#, when will it reach its maximum height??

1 Answer

Explanation:

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