A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #15 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

1 Answer
ali ergin
Mar 2, 2016

#"1) "x=v_i*t*cos alpha " horizontal movement"#
#"2) "y=v_i*t*sin alpha-1/2*g*t^2" vertical movement"#
#"3) "x~=5,734m#

Explanation:

enter image source here
#v_i:15 m/s " initial velocity"#
#alpha:(5cancel(pi))/12*180/cancel(pi)=75^o" sin75~=0,966 cos75~=0,259"#
#v_x=v_i*cos75 " x component of initial velocity"#
#v_y=v_i.sin75 " y component of initial velocity"#
#t_m=v_i*sin alpha/g "elapsed time to reach to max height"#
#h_m=(v_i^2*sin^2 alpha)/2*g" maximum height"#
#"1) "x=v_i*t*cos alpha " horizontal movement"#
#"2) "y=v_i*t*sin alpha-1/2*g*t^2" vertical movement"#
#"3) "t_m=(15*sin75)/(9,81~)=1,477 s#
#x=15*cos75*1,477#
#x~=5,734m#

Related questions
See all questions in Projectile Motion
Impact of this question
1780 views around the world
You can reuse this answer
Creative Commons License