A projectile is shot from the ground at an angle of #pi/12 # and a speed of #9 m/s#. When the projectile is at its maximum height, what will its distance from the starting point be?

1 Answer
Feb 4, 2016

#x=2,06422018349 ##meters#

Explanation:

#theta =cancel(pi)/12*180/cancel(pi)=15^o#
#" t is the elapsed time from the starting to maximum point"#
#t=v_i*sin 15/g#
#t=(9*0,258819045103)/(9,81)#
#t=0,237448665232 ##s#
#x=v_(ix)*t#
#v_(ix)=v_i*cos 15#
#v_(ix)=9*0,965925826289#
#v_(ix)=8,6933324366 ##m/s#
#x=8,6933324366*0,237448665232 #
#x=2,06422018349 ##meters#